Q: Q12. A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rot

Given:\nMass of the solid cylinder, M = 20 kg\nAngular speed, ω = 100 rad s⁻¹\nRadius of the cylinder, R = 0.25 m\n\n**Part 1: Kinetic energy associated with the rotation of the cylinder**\n\n**Step 1: Calculate the moment of inertia (I) of a solid cylinder about its axis.**\nThe formula for the moment of inertia of a solid cylinder about its own axis is I = (1/2)MR².\nI = (1/2) × 20 kg × (0.25 m)²\nI = 10 kg × 0.0625 m²\nI = 0.625 kg m²\n\n**Step 2: Calculate the rotational kinetic energy (K_rot).**\nThe formula for rotational kinetic energy is K_rot = (1/2)Iω².\nK_rot = (1/2) × 0.625 kg m² × (100 rad s⁻¹)²\nK_rot = (1/2) × 0.625 × 10000 J\nK_rot = 0.3125 × 10000 J\nK_rot = 3125 J\n\n**Therefore, the kinetic energy associated with the rotation of the cylinder is 3125 J.**\n\n--- \n\n**Part 2: Magnitude of angular momentum of the cylinder about its axis**\n\n**Step 1: Use the calculated moment of inertia (I).**\nFrom Part 1, I = 0.625 kg m².\n\n**Step 2: Calculate the angular momentum (L).**\nThe formula for angular momentum is L = Iω.\nL = 0.625 kg m² × 100 rad s⁻¹\nL = 62.5 kg m² s⁻¹\n\n**Therefore, the magnitude of the angular momentum of the cylinder about its axis is 62.5 kg m² s⁻¹.**

Question 1

Q1. Give the location of the centre of mass of a (a) sphere, (b) cylinder, (c) ring, and (d) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie insid

Accepted Answer

The location of the centre of mass (CM) for bodies of uniform mass density is determined by their geometric symmetry:

(a) **Sphere**: For a uniform sphere, the centre of mass is located at its geometric centre.

(b) **Cylinder**: For a uniform cylinder, the centre of mass is located at the midpoint of its axis of symmetry.

(c) **Ring**: For a uniform ring, the centre of mass is located at its geometric centre. This point lies in the empty space at the centre of the ring, not within the material of the ring itself.

(d) **Cube**: For a uniform cube, the centre of mass is located at its geometric centre, which is the point where its space diagonals intersect.

**Does the centre of mass of a body necessarily lie inside the body?**
No, the centre of mass of a body does not necessarily lie inside the body. As seen in the case of a uniform ring (part c), its centre of mass is at its geometric centre, which is an empty space. Other examples include a hollow sphere or a horseshoe, where the CM can be outside the physical material of the object.

Question 2

Q2. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10⁻¹⁰ m). Find the approximate location of the CM of the molecule, given that a chlor

Accepted Answer

The centre of mass (CM) of the HCl molecule can be found by treating it as a two-particle system. We will place the hydrogen atom at the origin and calculate the CM's position relative to it.

Question 3

Q3. A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child suddenly gets up and runs to the other end of the tr

Accepted Answer

Let's analyze the situation based on the fundamental principles of conservation of momentum and energy.

Question 4

Q4. Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

Accepted Answer

To show that the area of the triangle formed by two vectors `a` and `b` is half the magnitude of their cross product, we can use the geometric interpretation of the cross product.

Question 5

Q5. A uniform rod of length L and mass M is pivoted at one end. A spring of spring constant k is connected at the other end such that it is horizontal and the rod is in equilibrium in

Accepted Answer

To maintain the uniform rod in a horizontal equilibrium position, the spring must exert a vertical force that creates a torque balancing the torque due to the rod's weight. The weight of the uniform rod acts at its center of mass, which is at L/2 from the pivot. The spring is connected at the other end, which is at a distance L from the pivot. For rotational equilibrium, the net torque about the pivot must be zero.

Question 6

Q6. A non-uniform bar of weight W is suspended at two points P and Q. The distance between P and Q is L. The bar is in equilibrium. The forces exerted by the supports at P and Q are F_

Accepted Answer

For a non-uniform bar in equilibrium, both translational and rotational equilibrium conditions must be satisfied. We can use these conditions to find the position of its center of gravity (CG).

Question 7

Q7. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground

Accepted Answer

This problem involves the static equilibrium of the car. We need to apply both translational and rotational equilibrium conditions to find the normal forces exerted by the ground on the front and back wheels. Since the car is symmetrical, the force on each wheel of an axle will be half of the total force on that axle.

Question 8

Q8. Find the moment of inertia of a sphere about a tangent if the moment of inertia about its diameter is I.

Accepted Answer

To find the moment of inertia of a sphere about a tangent, we can use the Parallel Axis Theorem. The Parallel Axis Theorem states that if I_CM is the moment of inertia of a body about an axis passing through its center of mass, then the moment of inertia I about any other axis parallel to the first axis at a distance 'd' is given by I = I_CM + Md², where M is the mass of the body.

Question 9

Q9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground

Accepted Answer

Let M be the mass of the car, L be the distance between the front and back axles, and x be the distance of the center of gravity (CG) from the front axle.
Given:
Mass of the car, M = 1800 kg
Distance between axles, L = 1.8 m
Distance of CG from front axle, x = 1.05 m

We need to find the normal forces on the front wheels (N_f) and back wheels (N_b).

**Step 1: Calculate the total weight of the car.**
Weight (W) = Mg = 1800 kg × 9.8 m/s² = 17640 N

**Step 2: Apply the condition for translational equilibrium.**
Since the car is on a level ground and in equilibrium, the total upward normal force must balance the total downward weight.
N_f + N_b = W
N_f + N_b = 17640 N  --- (1)

**Step 3: Apply the condition for rotational equilibrium (sum of torques is zero).**
Let's take the front axle as the pivot point. The forces acting are:
1.  Weight (W) acting downwards at the CG, at a distance x from the front axle.
2.  Normal force on the back wheels (N_b) acting upwards at a distance L from the front axle.
3.  Normal force on the front wheels (N_f) passes through the pivot, so its torque is zero.

Torque due to weight (clockwise) = W × x
Torque due to back wheels (anticlockwise) = N_b × L

For rotational equilibrium:
N_b × L = W × x
N_b × 1.8 m = 17640 N × 1.05 m
N_b = (17640 × 1.05) / 1.8
N_b = 18522 / 1.8
N_b = 10290 N

**Step 4: Calculate the normal force on the front wheels.**
From equation (1):
N_f = W - N_b
N_f = 17640 N - 10290 N
N_f = 7350 N

**Step 5: Determine the force on each wheel.**
Assuming the weight is distributed equally between the two front wheels and two back wheels:
Force on each back wheel = N_b / 2 = 10290 N / 2 = 5145 N
Force on each front wheel = N_f / 2 = 7350 N / 2 = 3675 N

Therefore, the force exerted by the level ground on each front wheel is 3675 N, and on each back wheel is 5145 N.

Question 10

Q10. (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about its diameter to be (2/5)MR². (b) Given the moment of inert

Accepted Answer

**(a) Moment of inertia of a sphere about a tangent:**

**Given:**
Moment of inertia of a solid sphere about its diameter (I_CM) = (2/5)MR²

**Concept:** We can use the Parallel Axis Theorem.
The Parallel Axis Theorem states that I = I_CM + Md², where I is the moment of inertia about an axis parallel to the axis passing through the center of mass, I_CM is the moment of inertia about the axis passing through the center of mass, M is the mass of the body, and d is the perpendicular distance between the two parallel axes.

In this case, the tangent to the sphere is parallel to a diameter. The perpendicular distance (d) between the diameter (passing through the center of mass) and the tangent is equal to the radius (R) of the sphere.
So, d = R.

**Calculation:**
I_tangent = I_CM + MR²
I_tangent = (2/5)MR² + MR²
I_tangent = (2/5 + 1)MR²
I_tangent = (7/5)MR²

**Therefore, the moment of inertia of a sphere about a tangent to the sphere is (7/5)MR².**

---

**(b) Moment of inertia of a disc about a tangent in the plane of the disc:**

**Given:**
Moment of inertia of a disc about an axis normal to the disc and passing through its centre (I_z) = (1/2)MR²

**Concept:** This problem requires two steps:
1.  First, find the moment of inertia about a diameter using the Perpendicular Axis Theorem.
2.  Then, use the Parallel Axis Theorem to find the moment of inertia about a tangent in the plane of the disc.

**Step 1: Find the moment of inertia about a diameter (I_d) using the Perpendicular Axis Theorem.**
The Perpendicular Axis Theorem states that for a planar body, the moment of inertia about an axis perpendicular to its plane (I_z) is the sum of its moments of inertia about two perpendicular axes lying in the plane of the body (I_x and I_y), intersecting each other at the point where the perpendicular axis passes.
I_z = I_x + I_y

For a disc, due to symmetry, the moment of inertia about any diameter is the same. So, I_x = I_y = I_d.
Therefore, I_z = I_d + I_d = 2I_d
I_d = I_z / 2
I_d = (1/2)MR² / 2
I_d = (1/4)MR²

**Step 2: Find the moment of inertia about a tangent in the plane of the disc using the Parallel Axis Theorem.**
The tangent in the plane of the disc is parallel to a diameter. The perpendicular distance (d) between the diameter (passing through the center of mass) and this tangent is equal to the radius (R) of the disc.
So, d = R.

I_tangent_in_plane = I_d + MR²
I_tangent_in_plane = (1/4)MR² + MR²
I_tangent_in_plane = (1/4 + 1)MR²
I_tangent_in_plane = (5/4)MR²

**Therefore, the moment of inertia of a disc about a tangent in the plane of the disc is (5/4)MR².**

Question 11

Q11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. Both are free to rotate about their standard axes of symmetry.

Accepted Answer

To determine which body acquires a greater angular speed, we need to compare their angular accelerations, which in turn depend on their moments of inertia.

**1. Moments of Inertia:**
*   **Hollow Cylinder:** For a hollow cylinder rotating about its axis of symmetry, the moment of inertia (I_HC) is MR².
*   **Solid Sphere:** For a solid sphere rotating about its diameter (its axis of symmetry), the moment of inertia (I_SS) is (2/5)MR².

**Comparison of Moments of Inertia:**
Since (2/5) = 0.4, we have I_SS = 0.4 MR².
Clearly, I_HC = MR² > I_SS = (2/5)MR².
This means the hollow cylinder has a greater moment of inertia than the solid sphere for the same mass and radius.

**2. Angular Acceleration (α):**
We know that torque (τ) = Iα. Since the applied torques (τ) are of equal magnitude for both bodies:
α = τ / I

*   For the hollow cylinder: α_HC = τ / I_HC = τ / MR²
*   For the solid sphere: α_SS = τ / I_SS = τ / (2/5)MR² = (5/2)τ / MR²

**Comparison of Angular Accelerations:**
Since I_HC > I_SS, it implies that α_HC < α_SS. The solid sphere will have a greater angular acceleration.

**3. Angular Speed (ω) after a given time (t):**
Assuming both start from rest (ω₀ = 0), the angular speed after time t is given by the kinematic equation:
ω = ω₀ + αt
ω = αt

Since the time (t) is the same for both, the body with the greater angular acceleration (α) will acquire a greater angular speed (ω).

**Conclusion:**
As α_SS > α_HC, the **solid sphere** will acquire a greater angular speed after a given time.

**Reasoning Summary:** The solid sphere has a smaller moment of inertia compared to the hollow cylinder (for the same mass and radius). Since torque is equal to the product of moment of inertia and angular acceleration (τ = Iα), a smaller moment of inertia for the same applied torque results in a larger angular acceleration. A larger angular acceleration over the same time period leads to a greater final angular speed.

Question 12

Q12. A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rot

Accepted Answer

Given:
Mass of the solid cylinder, M = 20 kg
Angular speed, ω = 100 rad s⁻¹
Radius of the cylinder, R = 0.25 m

**Part 1: Kinetic energy associated with the rotation of the cylinder**

**Step 1: Calculate the moment of inertia (I) of a solid cylinder about its axis.**
The formula for the moment of inertia of a solid cylinder about its own axis is I = (1/2)MR².
I = (1/2) × 20 kg × (0.25 m)²
I = 10 kg × 0.0625 m²
I = 0.625 kg m²

**Step 2: Calculate the rotational kinetic energy (K_rot).**
The formula for rotational kinetic energy is K_rot = (1/2)Iω².
K_rot = (1/2) × 0.625 kg m² × (100 rad s⁻¹)²
K_rot = (1/2) × 0.625 × 10000 J
K_rot = 0.3125 × 10000 J
K_rot = 3125 J

**Therefore, the kinetic energy associated with the rotation of the cylinder is 3125 J.**

---

**Part 2: Magnitude of angular momentum of the cylinder about its axis**

**Step 1: Use the calculated moment of inertia (I).**
From Part 1, I = 0.625 kg m².

**Step 2: Calculate the angular momentum (L).**
The formula for angular momentum is L = Iω.
L = 0.625 kg m² × 100 rad s⁻¹
L = 62.5 kg m² s⁻¹

**Therefore, the magnitude of the angular momentum of the cylinder about its axis is 62.5 kg m² s⁻¹.**

Question 13

Q13. A uniform sphere of mass 20 kg and radius 0.25 m is rotating about an axis passing through its centre with an angular speed of 100 rad s–1. Calculate its (a) moment of inertia and

Accepted Answer

Given:
Mass of the uniform sphere, M = 20 kg
Radius of the uniform sphere, R = 0.25 m
Angular speed, ω = 100 rad s⁻¹

(a) Moment of inertia (I) of a uniform sphere about an axis passing through its centre is given by:
I = (2/5)MR²
I = (2/5) * 20 kg * (0.25 m)²
I = (2/5) * 20 * 0.0625
I = 8 * 0.0625
I = 0.5 kg m²

(b) Rotational kinetic energy (K_rot) is given by:
K_rot = (1/2)Iω²
K_rot = (1/2) * 0.5 kg m² * (100 rad s⁻¹)²
K_rot = (1/2) * 0.5 * 10000
K_rot = 0.25 * 10000
K_rot = 2500 J

Therefore, the moment of inertia of the sphere is 0.5 kg m² and its rotational kinetic energy is 2500 J.

Question 14

Q14. A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad s–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rot

Accepted Answer

Given:
Mass of the solid cylinder, M = 20 kg
Radius of the solid cylinder, R = 0.25 m
Angular speed, ω = 100 rad s⁻¹

(a) Kinetic energy associated with the rotation of the cylinder (Rotational Kinetic Energy):
First, calculate the moment of inertia (I) of a solid cylinder about its own axis.
I = (1/2)MR²
I = (1/2) * 20 kg * (0.25 m)²
I = 10 kg * 0.0625 m²
I = 0.625 kg m²

Now, calculate the rotational kinetic energy (K_rot):
K_rot = (1/2)Iω²
K_rot = (1/2) * 0.625 kg m² * (100 rad s⁻¹)²
K_rot = (1/2) * 0.625 * 10000
K_rot = 0.3125 * 10000
K_rot = 3125 J

(b) Magnitude of angular momentum (L) of the cylinder about its axis:
L = Iω
L = 0.625 kg m² * 100 rad s⁻¹
L = 62.5 kg m² s⁻¹

Therefore, the rotational kinetic energy of the cylinder is 3125 J and its angular momentum is 62.5 kg m² s⁻¹.

Question 15

Q15. A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the

Accepted Answer

Given:
Initial angular speed, ω₁ = 40 rev/min
Initial moment of inertia = I₁
Final moment of inertia, I₂ = (2/5)I₁

(a) Calculate the new angular speed:
First, convert the initial angular speed to rad/s:
ω₁ = 40 rev/min = 40 * (2π rad / 60 s) = (4π/3) rad/s

Since there is no external torque (friction is negligible), the angular momentum (L) of the child + turntable system is conserved.
L₁ = L₂
I₁ω₁ = I₂ω₂

Substitute I₂ = (2/5)I₁:
I₁ω₁ = (2/5)I₁ω₂
ω₁ = (2/5)ω₂
ω₂ = (5/2)ω₁

Now, substitute the value of ω₁:
ω₂ = (5/2) * (4π/3) rad/s
ω₂ = (10π/3) rad/s

To express in rev/min:
ω₂ = (10π/3) rad/s * (60 s / 2π rad) = (10 * 60) / (3 * 2) rev/min = 100 rev/min

So, the new angular speed of the child is (10π/3) rad/s or 100 rev/min.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation and account for the increase:
Initial rotational kinetic energy, K₁ = (1/2)I₁ω₁²
Final rotational kinetic energy, K₂ = (1/2)I₂ω₂²

Substitute I₂ = (2/5)I₁ and ω₂ = (5/2)ω₁:
K₂ = (1/2) * (2/5)I₁ * ((5/2)ω₁)²
K₂ = (1/2) * (2/5)I₁ * (25/4)ω₁²
K₂ = (1/2) * I₁ω₁² * (2/5) * (25/4)
K₂ = K₁ * (50/20)
K₂ = K₁ * (5/2)
K₂ = 2.5 K₁

Since K₂ = 2.5 K₁, the new kinetic energy of rotation is 2.5 times the initial kinetic energy of rotation, which means K₂ > K₁.

Account for the increase in kinetic energy:
The increase in kinetic energy comes from the internal work done by the child. When the child folds his arms, he pulls his arms inwards. This action requires muscular effort and involves doing positive work against the centrifugal force acting on his arms. This internal work done by the child is converted into the increased rotational kinetic energy of the system. The conservation of angular momentum holds because no external torque acts on the system, but the kinetic energy is not conserved because internal forces do work.

Question 16

Q16. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30

Accepted Answer

Given:
Mass of the hollow cylinder, M = 3 kg
Radius of the hollow cylinder, R = 40 cm = 0.40 m
Force applied to the rope, F = 30 N

(a) Angular acceleration (α) of the cylinder:
First, calculate the moment of inertia (I) of a hollow cylinder about its axis.
For a hollow cylinder, I = MR²
I = 3 kg * (0.40 m)²
I = 3 kg * 0.16 m²
I = 0.48 kg m²

The torque (τ) produced by the force F is given by:
τ = F * R
τ = 30 N * 0.40 m
τ = 12 N m

Now, apply Newton's second law for rotation: τ = Iα
α = τ / I
α = 12 N m / 0.48 kg m²
α = 25 rad s⁻²

(b) Linear acceleration (a) of the rope:
Since there is no slipping, the linear acceleration of the rope is related to the angular acceleration of the cylinder by:
a = Rα
a = 0.40 m * 25 rad s⁻²
a = 10 m s⁻²

Therefore, the angular acceleration of the cylinder is 25 rad s⁻² and the linear acceleration of the rope is 10 m s⁻².

Question 17

Q17. To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: Uniform angular vel

Accepted Answer

Given:
Uniform angular speed, ω = 200 rad s⁻¹
Torque transmitted by the engine, τ = 180 N m

The power (P) required by the engine in rotational motion is given by the formula:
P = τω

Substitute the given values:
P = 180 N m * 200 rad s⁻¹
P = 36000 W

To express in kilowatts (kW):
P = 36000 W / 1000 W/kW = 36 kW

Therefore, the power required by the engine is 36000 W or 36 kW.

Question 18

Q18. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.
(a) How far w

Accepted Answer

The cylinder rolls up an inclined plane. We will use the principle of conservation of mechanical energy to find the maximum height reached and then use kinematic equations to find the time taken to return.

**(a) Distance travelled up the plane:**
Given:
Angle of inclination, θ = 30°
Initial speed of centre of mass, v = 5 m/s

For a solid cylinder, the moment of inertia about its axis is I = (1/2)MR².
When the cylinder rolls without slipping, v = Rω.

Initial total mechanical energy at the bottom of the inclined plane (h=0):
E_initial = Translational KE + Rotational KE + Potential Energy
E_initial = (1/2)Mv² + (1/2)Iω² + 0
Substitute I = (1/2)MR² and ω = v/R:
E_initial = (1/2)Mv² + (1/2)((1/2)MR²)(v/R)²
E_initial = (1/2)Mv² + (1/4)Mv²
E_initial = (3/4)Mv²

At the maximum height 'h' reached up the plane, the cylinder momentarily comes to rest, so its final velocity (translational and rotational) is zero.
E_final = Potential Energy at height h
E_final = Mgh

By conservation of mechanical energy:
E_initial = E_final
(3/4)Mv² = Mgh
h = (3/4)v²/g

Substitute the given values:
h = (3/4) * (5 m/s)² / (9.8 m/s²)
h = (3/4) * 25 / 9.8
h = 75 / (4 * 9.8)
h = 75 / 39.2
h ≈ 1.913 m

The distance 's' travelled up the inclined plane is related to the height 'h' by sinθ = h/s.
s = h / sinθ
s = 1.913 m / sin(30°)
s = 1.913 m / 0.5
s = 3.826 m

So, the cylinder will go approximately **3.83 m** up the plane.

**(b) Time taken to return to the bottom:**
First, let's find the acceleration of the cylinder rolling down the inclined plane.
The forces acting on the cylinder are:
1. Gravitational force, Mg, acting vertically downwards.
2. Normal force, N, perpendicular to the plane.
3. Frictional force, f, acting up the plane (to cause rolling without slipping).

Equations of motion:
Along the inclined plane: Mg sinθ - f = Ma_cm  (1)
Perpendicular to the inclined plane: N - Mg cosθ = 0
Torque equation about the centre of mass: fR = Iα
Since rolling without slipping, a_cm = Rα, so α = a_cm/R.
fR = I(a_cm/R)
f = (I/R²)a_cm
For a solid cylinder, I = (1/2)MR².
f = ((1/2)MR²/R²)a_cm
f = (1/2)Ma_cm  (2)

Substitute (2) into (1):
Mg sinθ - (1/2)Ma_cm = Ma_cm
Mg sinθ = (3/2)Ma_cm
a_cm = (2/3)g sinθ

Substitute the values:
a_cm = (2/3) * 9.8 m/s² * sin(30°)
a_cm = (2/3) * 9.8 * 0.5
a_cm = (1/3) * 9.8
a_cm ≈ 3.267 m/s²

Now, we use kinematic equations for motion down the incline.
Initial velocity at the top (max height) is u = 0.
Distance to travel is s = 3.826 m.
Acceleration is a_cm = 3.267 m/s².
Let 't' be the time taken to return to the bottom.
Using the equation: s = ut + (1/2)at²
3.826 = 0 * t + (1/2) * 3.267 * t²
3.826 = (1/2) * 3.267 * t²
t² = (2 * 3.826) / 3.267
t² = 7.652 / 3.267
t² ≈ 2.342
t = √2.342
t ≈ 1.530 s

So, it will take approximately **1.53 s** to return to the bottom.

Question 19

Q19. A disc of radius R is cut from a uniform thin metallic disc of radius 2R. Its centre is at (0, 0). The centre of the hole is at (R, 0). Find the coordinates of the centre of mass o

Accepted Answer

To find the centre of mass of the remaining part of the disc, we can use the concept of negative mass or the subtraction method. We consider the original large disc as a complete system and subtract the properties of the cut-out disc (hole).

Let σ be the surface mass density (mass per unit area) of the uniform metallic disc.

**1. Original Disc (before cutting):**
Radius = 2R
Area, A_original = π(2R)² = 4πR²
Mass, M_original = A_original * σ = 4πR²σ
Centre of mass, X_original = 0, Y_original = 0 (since its centre is at (0,0))

**2. Cut-out Disc (the hole):**
Radius = R
Area, A_hole = πR²
Mass, M_hole = A_hole * σ = πR²σ
Centre of mass, X_hole = R, Y_hole = 0 (since the centre of the hole is at (R,0))

**3. Remaining Part:**
Mass, M_remaining = M_original - M_hole = 4πR²σ - πR²σ = 3πR²σ

Now, we use the formula for the centre of mass of a system with a hole:
X_CM = (M_original * X_original - M_hole * X_hole) / (M_original - M_hole)
Y_CM = (M_original * Y_original - M_hole * Y_hole) / (M_original - M_hole)

**Calculate X_CM:**
X_CM = ( (4πR²σ) * 0 - (πR²σ) * R ) / (3πR²σ)
X_CM = ( 0 - πR³σ ) / (3πR²σ)
X_CM = -πR³σ / (3πR²σ)
X_CM = -R/3

**Calculate Y_CM:**
Y_CM = ( (4πR²σ) * 0 - (πR²σ) * 0 ) / (3πR²σ)
Y_CM = 0 / (3πR²σ)
Y_CM = 0

Therefore, the coordinates of the centre of mass of the remaining part of the disc are **(-R/3, 0)**.

Question 20

Q20. A solid sphere rolls without slipping on an inclined plane and reaches the bottom with velocity v. If the same sphere slides without friction on the same inclined plane and reaches

Accepted Answer

Let the mass of the solid sphere be M, its radius be R, and the height of the inclined plane be h.

**Case 1: Solid sphere rolls without slipping.**
When the solid sphere rolls without slipping, its initial potential energy (Mgh) at the top of the inclined plane is converted into both translational kinetic energy and rotational kinetic energy at the bottom.

By conservation of mechanical energy:
Initial Potential Energy = Final Translational KE + Final Rotational KE
Mgh = (1/2)Mv² + (1/2)Iω²

For a solid sphere, the moment of inertia I = (2/5)MR².
For rolling without slipping, v = Rω, so ω = v/R.

Substitute I and ω into the energy equation:
Mgh = (1/2)Mv² + (1/2)((2/5)MR²)(v/R)²
Mgh = (1/2)Mv² + (1/5)Mv²
Mgh = ( (5+2)/10 )Mv²
Mgh = (7/10)Mv²
v² = (10/7)gh
v = √( (10/7)gh )

**Case 2: Solid sphere slides without friction.**
When the solid sphere slides without friction, there is no rotational motion, and thus no rotational kinetic energy. Its initial potential energy (Mgh) is converted entirely into translational kinetic energy at the bottom.

By conservation of mechanical energy:
Initial Potential Energy = Final Translational KE
Mgh = (1/2)M(v')²
(v')² = 2gh
v' = √(2gh)

**Comparison:**
We need to compare v = √( (10/7)gh ) and v' = √(2gh).

Let's compare the coefficients of gh:
For v: 10/7 ≈ 1.428
For v': 2

Since 2 > 10/7, it means v' > v.

Therefore, the velocity **v' (when sliding without friction) is greater** than the velocity v (when rolling without slipping).

**Reason:**
When the sphere rolls without slipping, a part of its initial potential energy is converted into rotational kinetic energy, and the remaining part is converted into translational kinetic energy. In contrast, when the sphere slides without friction, all of its initial potential energy is converted solely into translational kinetic energy. Since the total energy available (Mgh) is the same in both cases, the translational kinetic energy (and hence the translational velocity) will be greater when no energy is 'diverted' into rotational motion.

Question 21

Q21. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation

Accepted Answer

Given:
Mass of the solid cylinder, M = 20 kg
Angular speed, ω = 100 rad s⁻¹
Radius of the cylinder, R = 0.25 m

**1. Moment of Inertia (I) of a solid cylinder:**
The moment of inertia of a solid cylinder rotating about its own axis is given by:
I = (1/2)MR²

Substitute the given values:
I = (1/2) * 20 kg * (0.25 m)²
I = 10 kg * 0.0625 m²
I = 0.625 kg m²

**2. Kinetic Energy associated with rotation (KE_rot):**
The rotational kinetic energy is given by:
KE_rot = (1/2)Iω²

Substitute the calculated I and given ω:
KE_rot = (1/2) * 0.625 kg m² * (100 rad s⁻¹)²
KE_rot = (1/2) * 0.625 * 10000 J
KE_rot = 0.3125 * 10000 J
KE_rot = 3125 J

**3. Magnitude of Angular Momentum (L):**
The angular momentum of the cylinder about its axis is given by:
L = Iω

Substitute the calculated I and given ω:
L = 0.625 kg m² * 100 rad s⁻¹
L = 62.5 kg m² s⁻¹

Therefore:
Kinetic energy associated with the rotation of the cylinder is **3125 J**.
The magnitude of angular momentum of the cylinder about its axis is **62.5 kg m² s⁻¹**.

Question 22

Q22. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b)

Accepted Answer

(a) Yes, the solid sphere will reach the bottom with the same speed in each case.

(b) Yes, it will take longer to roll down one plane than the other.

(c) The sphere will take longer to roll down the inclined plane with the smaller angle of inclination. This is because the acceleration of the sphere down the incline is directly proportional to sinθ, where θ is the angle of inclination. A smaller θ means a smaller acceleration, and thus a longer time to cover the same vertical height (or equivalent distance along the incline, considering the net effect of acceleration).

Question 23

Q23. A disc of radius R is pulled by a force F acting at its rim. The force is parallel to the plane of the disc and acts perpendicular to the radius. What is the torque produced by the

Accepted Answer

The torque produced by the force about the center of the disc is FR.

Question 24

Q24. A solid cylinder of mass 20 kg and radius 0.25 m is rotating about its own axis with an angular speed of 100 rad s⁻¹. What is the kinetic energy associated with the rotation of the

Accepted Answer

The kinetic energy associated with the rotation of the cylinder is 625 J.
The magnitude of the angular momentum of the cylinder about its axis is 12.5 kg m² s⁻¹.

Question 25

Q25. A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the

Accepted Answer

(a) The new angular speed of the turntable is 100 rev/min.
(b) The child's new rotational kinetic energy is 2.5 times the initial rotational kinetic energy. This increase in kinetic energy is accounted for by the work done by the child in pulling his arms inwards against the centrifugal force. This internal work done by the child is converted into rotational kinetic energy of the system.

Question 26

Q26. A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30

Accepted Answer

The angular acceleration of the cylinder is 62.5 rad s⁻².
The linear acceleration of the rope is 25 m s⁻².

Question 27

Q27. A disc rotating at 33 1/3 rev/min is brought to rest in 20 s. Calculate the angular acceleration and the number of revolutions made before coming to rest.

Accepted Answer

The initial angular speed of the disc is 33 1/3 rev/min, which is 100/3 rev/min. This needs to be converted to rad/s. The final angular speed is 0 rad/s as it comes to rest. The time taken is 20 s.

First, convert the initial angular speed to rad/s:
ω₀ = (100/3) rev/min = (100/3) * (2π rad / 60 s) = (100 * 2π) / 180 rad/s = (10π/9) rad/s.

Using the first equation of rotational kinematics: ω = ω₀ + αt
0 = (10π/9) + α(20)
α = -(10π/9) / 20 = -π/18 rad/s².

The negative sign indicates that the angular acceleration is opposite to the direction of initial angular velocity, meaning it's a deceleration.

Next, calculate the angular displacement (θ) using the equation: θ = ω₀t + (1/2)αt²
θ = (10π/9)(20) + (1/2)(-π/18)(20)²
θ = (200π/9) - (π/36)(400)
θ = (200π/9) - (100π/9)
θ = 100π/9 radians.

To find the number of revolutions, convert radians to revolutions:
Number of revolutions = θ / (2π)
Number of revolutions = (100π/9) / (2π) = 100/18 = 50/9 revolutions.
Number of revolutions ≈ 5.56 revolutions.

Alternatively, using ω² = ω₀² + 2αθ:
0² = (10π/9)² + 2(-π/18)θ
0 = (100π²/81) - (π/9)θ
(π/9)θ = 100π²/81
θ = (100π²/81) * (9/π) = 100π/9 radians.
This confirms the angular displacement.

Number of revolutions = (100π/9) / (2π) = 50/9 ≈ 5.56 revolutions.

**Final Answer:**
Angular acceleration (α) = -π/18 rad/s² ≈ -0.1745 rad/s²
Number of revolutions = 50/9 revolutions ≈ 5.56 revolutions

Question 28

Q28. A wheel starting from rest, rotates with a constant angular acceleration of 4.0 rad s⁻². What is its angular velocity after 8.0 s? How much angular displacement does it cover in th

Accepted Answer

Given:
Initial angular velocity, ω₀ = 0 rad/s (starting from rest)
Angular acceleration, α = 4.0 rad s⁻²
Time, t = 8.0 s

**Part 1: Calculate angular velocity after 8.0 s**
Using the first equation of rotational kinematics:
ω = ω₀ + αt
ω = 0 + (4.0 rad s⁻²) * (8.0 s)
ω = 32.0 rad s⁻¹

**Part 2: Calculate angular displacement in this time**
Using the second equation of rotational kinematics:
θ = ω₀t + (1/2)αt²
θ = (0 rad/s) * (8.0 s) + (1/2) * (4.0 rad s⁻²) * (8.0 s)²
θ = 0 + (1/2) * (4.0) * (64.0)
θ = 2.0 * 64.0
θ = 128.0 radians

**Final Answer:**
Angular velocity after 8.0 s = 32.0 rad s⁻¹
Angular displacement in 8.0 s = 128.0 radians

Question 29

Q29. A uniform rod of mass 20 kg and length 2 m is pivoted at one end and allowed to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angu

Accepted Answer

This problem can be solved using the principle of conservation of mechanical energy. As the rod rotates from a horizontal to a vertical position, its potential energy decreases, and this decrease is converted into rotational kinetic energy.

**1. Identify Given Parameters:**
Mass of the rod, M = 20 kg
Length of the rod, L = 2 m
Initial position: Horizontal (released from rest)
Final position: Vertical

**2. Calculate Moment of Inertia (I):**
The rod is pivoted at one end. For a uniform rod of mass M and length L, pivoted at one end, its moment of inertia is given by:
I = (1/3)ML²
I = (1/3) * (20 kg) * (2 m)²
I = (1/3) * 20 * 4
I = 80/3 kg m²

**3. Calculate Change in Potential Energy (ΔPE):**
When the rod is in the horizontal position, its center of mass (CM) is at a height L/2 above the pivot (if we consider the pivot as the reference level for potential energy, and the rod is initially horizontal, the CM is at the same height as the pivot). However, it's more convenient to consider the initial horizontal position as the reference (h=0) for the CM. When the rod rotates to the vertical position, its center of mass moves downwards by a distance equal to L/2.

Initial height of CM (h_initial) = 0 (taking the horizontal position of CM as reference)
Final height of CM (h_final) = -L/2 (below the initial position)

Change in potential energy, ΔPE = PE_final - PE_initial
ΔPE = Mg(h_final) - Mg(h_initial)
ΔPE = Mg(-L/2) - 0
ΔPE = -MgL/2

ΔPE = -(20 kg) * (9.8 m/s²) * (2 m / 2)
ΔPE = -20 * 9.8 * 1
ΔPE = -196 J

The decrease in potential energy is |ΔPE| = 196 J.

**4. Apply Conservation of Mechanical Energy:**
Initial mechanical energy (E_initial) = Final mechanical energy (E_final)
E_initial = PE_initial + KE_initial
E_final = PE_final + KE_final

Since the rod is released from rest, initial kinetic energy (KE_initial) = 0.
Initial potential energy (PE_initial) = 0 (by choosing reference).

So, 0 + 0 = PE_final + KE_final
0 = -MgL/2 + (1/2)Iω²

Where ω is the angular speed when the rod reaches the vertical position.

(1/2)Iω² = MgL/2
Iω² = MgL

Substitute the value of I:
(80/3)ω² = (20 kg) * (9.8 m/s²) * (2 m)
(80/3)ω² = 392

ω² = (392 * 3) / 80
ω² = 1176 / 80
ω² = 14.7

ω = √14.7
ω ≈ 3.834 rad/s

**Final Answer:**
The angular speed of the rod when it reaches the vertical position is approximately 3.83 rad/s.

Question 30

Q30. A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad s⁻¹. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rot

Accepted Answer

Given:
Mass of the solid cylinder, M = 20 kg
Angular speed, ω = 100 rad s⁻¹
Radius of the cylinder, R = 0.25 m

**Part 1: Calculate the moment of inertia (I) of the solid cylinder.**
For a solid cylinder rotating about its own axis, the moment of inertia is given by:
I = (1/2)MR²
I = (1/2) * (20 kg) * (0.25 m)²
I = 10 kg * (0.0625 m²)
I = 0.625 kg m²

**Part 2: Calculate the kinetic energy associated with the rotation (KE_rot).**
The rotational kinetic energy is given by:
KE_rot = (1/2)Iω²
KE_rot = (1/2) * (0.625 kg m²) * (100 rad s⁻¹)²
KE_rot = (1/2) * 0.625 * 10000
KE_rot = 0.3125 * 10000
KE_rot = 3125 J

**Part 3: Calculate the magnitude of the angular momentum (L) of the cylinder.**
The angular momentum about the axis of rotation is given by:
L = Iω
L = (0.625 kg m²) * (100 rad s⁻¹)
L = 62.5 kg m² s⁻¹

**Final Answer:**
Kinetic energy associated with the rotation = 3125 J
Magnitude of the angular momentum = 62.5 kg m² s⁻¹

Question 31

Q31. Find the torque of a force 7i + 3j - 5k about the origin. The force acts on a particle whose position vector is i - j + k.

Accepted Answer

The torque (τ) of a force (F) about the origin, when the force acts on a particle with position vector (r), is given by the cross product τ = r × F.

Given:
Position vector, r = i - j + k
Force vector, F = 7i + 3j - 5k

Calculating the cross product:
τ = r × F = (i - j + k) × (7i + 3j - 5k)

Using the determinant method:
τ = | i   j   k |
    | 1  -1   1 |
    | 7   3  -5 |

τ = i [(-1)(-5) - (1)(3)] - j [(1)(-5) - (1)(7)] + k [(1)(3) - (-1)(7)]
τ = i [5 - 3] - j [-5 - 7] + k [3 + 7]
τ = i [2] - j [-12] + k [10]
τ = 2i + 12j + 10k

The torque about the origin is 2i + 12j + 10k N m.

Question 32

Q32. Show that the angular momentum about any point of a single particle moving with constant velocity remains constant.

Accepted Answer

### Introduction
Angular momentum (L) of a particle about a point is defined as the cross product of its position vector (r) relative to that point and its linear momentum (p). We need to show that if a particle moves with constant velocity, its angular momentum about any point remains constant.

### Derivation
1.  **Definition of Angular Momentum:**
    The angular momentum L of a particle with position vector r and linear momentum p is given by:
    L = r × p

2.  **Linear Momentum in terms of Mass and Velocity:**
    The linear momentum p of a particle of mass m moving with velocity v is:
    p = mv
    So, L = r × (mv)

3.  **Rate of Change of Angular Momentum:**
    To show that L is constant, we need to prove that its time derivative (dL/dt) is zero. Differentiating L with respect to time:
    dL/dt = d/dt (r × p)

4.  **Applying the Product Rule for Cross Products:**
    Using the product rule for differentiation of cross products (similar to (uv)' = u'v + uv'):
    dL/dt = (dr/dt) × p + r × (dp/dt)

5.  **Analyzing the First Term (dr/dt × p):**
    *   dr/dt is the velocity (v) of the particle.
    *   So, dr/dt × p = v × p
    *   Substitute p = mv:
        v × p = v × (mv) = m (v × v)
    *   The cross product of a vector with itself is always zero (v × v = 0).
    *   Therefore, (dr/dt) × p = 0.

6.  **Analyzing the Second Term (r × dp/dt):**
    *   dp/dt is the rate of change of linear momentum, which, according to Newton's second law, is equal to the net force (F) acting on the particle.
    *   So, r × (dp/dt) = r × F
    *   The term r × F is the definition of torque (τ) acting on the particle about the chosen point.
    *   Therefore, r × (dp/dt) = τ.

7.  **Relating dL/dt to Torque:**
    Combining the results from steps 5 and 6:
    dL/dt = 0 + τ
    dL/dt = τ
    This is the rotational analogue of Newton's second law.

8.  **Condition for Constant Velocity:**
    The problem states that the particle is moving with constant velocity. If the velocity (v) is constant, then:
    *   The acceleration (a) of the particle is zero (a = dv/dt = 0).
    *   According to Newton's second law (F = ma), the net force (F) acting on the particle must be zero (F = m * 0 = 0).

9.  **Conclusion:**
    Since the net force F = 0, the torque τ = r × F must also be zero (τ = r × 0 = 0).
    Substituting τ = 0 into the equation dL/dt = τ:
    dL/dt = 0

If the rate of change of angular momentum is zero, it means that the angular momentum L is constant.

### Summary
For a single particle moving with constant velocity, the net force acting on it is zero. Consequently, the torque acting on it about any point is also zero. Since the rate of change of angular momentum is equal to the net torque, a zero torque implies a constant angular momentum. Thus, the angular momentum about any point of a single particle moving with constant velocity remains constant.

Question 33

Q33. A uniform rod of mass m and length l is pivoted at one end A. It is held in a horizontal position and released from rest. When the rod is vertical, find the speed of its centre of

Accepted Answer

### Problem Description
A uniform rod of mass 'm' and length 'l' is pivoted at one end (A). It is initially held horizontally and released from rest. We need to find the speed of its centre of mass (v_CM) and its angular speed (ω) when it reaches the vertical position.

### Approach
We will use the principle of conservation of mechanical energy, as only conservative forces (gravity) are doing work. The pivot force does no work because the point of application is stationary.

#### 1. Initial State (Horizontal Position)
*   **Potential Energy (PE_i):** Let's set the reference level for potential energy at the pivot point A. The centre of mass (CM) of a uniform rod is at its geometric center, i.e., at a distance l/2 from the pivot. In the horizontal position, the CM is at the same vertical level as the pivot. Therefore, the initial potential energy is PE_i = 0.
*   **Kinetic Energy (KE_i):** The rod is released from rest, so its initial kinetic energy is KE_i = 0.
*   **Total Initial Energy (E_i):** E_i = PE_i + KE_i = 0 + 0 = 0.

#### 2. Final State (Vertical Position)
*   **Potential Energy (PE_f):** When the rod is vertical, its centre of mass (CM) has moved downwards by a distance equal to l/2 from its initial horizontal position. Since we set the reference at the pivot, and the CM is now below the pivot, the final potential energy is negative.
    PE_f = -mg(l/2)
*   **Kinetic Energy (KE_f):** In the vertical position, the rod is rotating about the pivot A. Its kinetic energy is purely rotational.
    KE_f = (1/2)Iω²
    Where I is the moment of inertia of the rod about the pivot A, and ω is the angular speed.
    For a uniform rod of mass m and length l pivoted at one end, the moment of inertia is I = (1/3)ml².
    So, KE_f = (1/2) * (1/3)ml² * ω² = (1/6)ml²ω²
*   **Total Final Energy (E_f):** E_f = PE_f + KE_f = -mg(l/2) + (1/6)ml²ω²

#### 3. Conservation of Mechanical Energy
According to the principle of conservation of mechanical energy:
E_i = E_f
0 = -mg(l/2) + (1/6)ml²ω²

Rearrange the equation to solve for ω²:
mg(l/2) = (1/6)ml²ω²

Cancel 'm' and one 'l' from both sides:
g/2 = (1/6)lω²

Multiply by 6:
3g = lω²

Solve for ω:
ω² = 3g/l
ω = √(3g/l)

This is the angular speed of the rod when it is vertical.

#### 4. Speed of the Centre of Mass (v_CM)
The linear speed of the centre of mass (v_CM) is related to the angular speed (ω) by:
v_CM = r_CM * ω
Where r_CM is the distance of the centre of mass from the pivot. For a uniform rod, r_CM = l/2.

v_CM = (l/2) * ω
Substitute the value of ω:
v_CM = (l/2) * √(3g/l)
v_CM = (l/2) * (√3 * √g / √l)
v_CM = (√l * √l / 2) * (√3 * √g / √l)
v_CM = (√3 * √g * √l) / 2
v_CM = (1/2)√(3gl)

### Final Answer
*   **Angular speed of the rod (ω):** √(3g/l)
*   **Speed of its centre of mass (v_CM):** (1/2)√(3gl)

Chapter 7: System of Particles and Rotational Motion