Question 1

1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?

Accepted Answer

For each of the given bodies with uniform mass density, the centre of mass (CM) is located at the geometric centre due to symmetry:

(i) Sphere: The CM is at the centre of the sphere.

(ii) Cylinder: The CM is at the midpoint of the axis of the cylinder, i.e., at the centre along its length and at the centre of the circular cross-section.

(iii) Ring: The CM is at the centre of the ring (the centre of the circle formed by the ring).

(iv) Cube: The CM is at the geometric centre of the cube.

Regarding whether the CM necessarily lies inside the body: No, the CM does not necessarily lie inside the body. For example, in a ring or a hollow object, the CM can lie in the empty space inside the ring or hollow part, which is outside the material of the body. — The centre of mass for uniform bodies coincides with their geometric centre due to uniform distribution of mass and symmetry. However, for bodies with holes or non-convex shapes, the CM can lie outside the material body itself.

Question 2

2 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 chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Accepted Answer

Let the mass of hydrogen atom = m
Mass of chlorine atom = 35.5 m
Distance between nuclei = d = 1.27 × 10^{-10} m

Taking hydrogen at origin (x=0), chlorine at x=d,

Centre of mass (CM) position from hydrogen nucleus:

x_CM = (m × 0 + 35.5 m × d) / (m + 35.5 m) = (35.5 d) / 36.5

x_CM = (35.5 / 36.5) × 1.27 × 10^{-10} m ≈ 1.236 × 10^{-10} m

So, the CM lies approximately 1.236 Å from the hydrogen nucleus towards chlorine. — Using the definition of centre of mass for two particles along a line:

x_CM = (m1 x1 + m2 x2) / (m1 + m2)

Substituting values and simplifying gives the position of CM closer to the heavier chlorine atom.

Question 3

3 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 gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?

Accepted Answer

The speed of the centre of mass (CM) of the system (trolley + child) remains the same as the initial speed V of the trolley.

Explanation:
Since there are no external horizontal forces acting on the system (assuming smooth floor and no friction), the velocity of the CM cannot change due to internal motions of the child on the trolley. The child running inside the trolley is an internal motion and does not affect the CM velocity.

Therefore, the speed of the CM of the system remains V. — The velocity of the CM of a system is governed by external forces. Internal motions do not change the CM velocity. Hence, the CM speed remains constant at V.

Question 4

4 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

The area of the parallelogram formed by vectors a and b is |a × b|.

Since a triangle is half of a parallelogram, the area of the triangle formed by vectors a and b is:

Area = (1/2) |a × b|

Proof:
The magnitude of the cross product |a × b| = |a||b| sin θ, where θ is the angle between a and b.

The area of the parallelogram with sides a and b is base × height = |a||b| sin θ = |a × b|.

Therefore, the area of the triangle = (1/2) × area of parallelogram = (1/2) |a × b|. — Using the geometric interpretation of the cross product, the magnitude gives the area of parallelogram formed by two vectors. Triangle area is half of that.

Question 5

5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.

Accepted Answer

The scalar triple product a · (b × c) gives the volume of the parallelepiped formed by vectors a, b, and c.

Proof:
- The vector b × c is perpendicular to the plane containing b and c, and its magnitude is the area of the parallelogram formed by b and c.
- The scalar product a · (b × c) projects vector a along the direction perpendicular to b and c.

Therefore, the magnitude |a · (b × c)| = |a| × (area of parallelogram formed by b and c) × cos θ, where θ is the angle between a and (b × c).

Since (b × c) is perpendicular to b and c, the angle θ is the angle between a and the normal to the base parallelogram, so the product gives the height of the parallelepiped.

Hence, volume = base area × height = |a · (b × c)|. — The scalar triple product geometrically represents the volume of the parallelepiped formed by three vectors.

Question 6

6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components p_x, p_y and p_z. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Accepted Answer

Angular momentum l = r × p

Given:
r = (x, y, z)
p = (p_x, p_y, p_z)

Components of l:

l_x = y p_z - z p_y
l_y = z p_x - x p_z
l_z = x p_y - y p_x

If the particle moves only in the x-y plane, then z = 0 and p_z = 0.

Substituting z = 0 and p_z = 0:

l_x = y × 0 - 0 × p_y = 0
l_y = 0 × p_x - x × 0 = 0
l_z = x p_y - y p_x (non-zero in general)

Thus, the angular momentum vector has only the z-component when the motion is confined to the x-y plane. — Using the definition of angular momentum as cross product of position and momentum vectors, components are derived. Restricting motion to x-y plane eliminates x and y components of angular momentum.

Question 7

7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

Accepted Answer

Let the two particles be moving along lines parallel to the x-axis, separated by distance d along y-axis.

Particle 1: position vector r_1 = (0, 0, 0), velocity v along +x
Particle 2: position vector r_2 = (0, d, 0), velocity v along -x

Angular momentum of particle 1 about point O:
L_1 = r_1 × p_1 = 0 (since r_1 = 0)

Angular momentum of particle 2 about point O:
L_2 = r_2 × p_2 = (0, d, 0) × (-m v, 0, 0) = (0, d, 0) × (-m v, 0, 0) = (0, 0, m v d)

Total angular momentum about O:
L = L_1 + L_2 = (0, 0, m v d)

Now, consider another point O' displaced by vector R. The angular momentum about O' is:

L' = Σ (r_i - R) × p_i = Σ r_i × p_i - R × Σ p_i

Since the two particles have equal and opposite momenta, total momentum Σ p_i = 0.

Therefore, L' = L

Hence, the angular momentum vector of the system is independent of the choice of origin. — The total angular momentum about any point differs by R × total momentum. Since total momentum is zero, angular momentum is invariant with respect to origin.

Question 8

8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Accepted Answer

Given:
Weight of bar = W
Length of bar = L = 2 m
Angles with vertical: θ1 = 36.9°, θ2 = 53.1°

Let T1 and T2 be the tensions in the two strings.

Resolve forces vertically:
T1 cos θ1 + T2 cos θ2 = W  ...(1)

Resolve forces horizontally:
T1 sin θ1 = T2 sin θ2  ...(2)

From (2):
T1 = T2 (sin θ2 / sin θ1)

Substitute in (1):
T2 (sin θ2 / sin θ1) cos θ1 + T2 cos θ2 = W

Calculate sin and cos values:
sin 36.9° ≈ 0.6, cos 36.9° ≈ 0.8
sin 53.1° ≈ 0.8, cos 53.1° ≈ 0.6

T2 (0.8 / 0.6) × 0.8 + T2 × 0.6 = W
T2 ( (0.8 × 0.8)/0.6 + 0.6 ) = W
T2 (1.0667 + 0.6) = W
T2 × 1.6667 = W
T2 = W / 1.6667

T1 = T2 (0.8 / 0.6) = (W / 1.6667) × 1.3333 = 0.8 W

Taking moments about left end:
T1 × 0 + T2 × L cos θ2 = W × d

Note: The horizontal distances of the tensions from the left end are zero for T1 (left end) and L cos θ2 for T2.

Calculate L cos θ2 = 2 × 0.6 = 1.2 m

Moment balance:
T2 × 1.2 = W × d

Substitute T2:
(W / 1.6667) × 1.2 = W × d

Simplify:
(1.2 / 1.6667) W = W d

d = 0.72 m

Therefore, the centre of gravity is 0.72 m from the left end. — Using equilibrium of forces and moments, tensions are found and then moment balance gives the location of centre of gravity.

Question 9

9 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 on each front wheel and each back wheel.

Accepted Answer

Given:
Mass of car, m = 1800 kg
Weight, W = mg = 1800 × 9.8 = 17640 N
Distance between axles, L = 1.8 m
Distance of centre of gravity from front axle, d = 1.05 m

Let R_f = reaction force on front wheels
R_b = reaction force on back wheels

Sum of vertical forces:
R_f + R_b = W = 17640 N ...(1)

Taking moments about front axle:
R_b × L = W × d
R_b × 1.8 = 17640 × 1.05
R_b = (17640 × 1.05) / 1.8 = 10290 N

From (1):
R_f = 17640 - 10290 = 7349.9 N

Therefore,
Force on front wheels = 7349.9 N
Force on back wheels = 10290 N — Using equilibrium of forces and moments, reactions on front and back wheels are calculated.

Question 10

10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.

Accepted Answer

Given equal torques τ applied to both bodies for the same time interval.

Moment of inertia of hollow cylinder about its axis:
I_cylinder = M R^2

Moment of inertia of solid sphere about its axis:
I_sphere = (2/5) M R^2

Angular acceleration α = τ / I

Since I_sphere < I_cylinder, the sphere will have a greater angular acceleration.

For the same time t, angular speed ω = α t

Therefore, the solid sphere will acquire a greater angular speed after the given time. — Angular acceleration is inversely proportional to moment of inertia. Smaller moment of inertia leads to higher angular acceleration and thus greater angular speed for same torque and time.

Question 11

11 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 of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Accepted Answer

Given:
Mass, M = 20 kg
Radius, R = 0.25 m
Angular speed, ω = 100 rad/s

Moment of inertia of solid cylinder about its axis:
I = (1/2) M R^2 = (1/2) × 20 × (0.25)^2 = 0.625 kg·m^2

Kinetic energy of rotation:
K = (1/2) I ω^2 = (1/2) × 0.625 × (100)^2 = 0.3125 × 10000 = 3125 J

Angular momentum:
L = I ω = 0.625 × 100 = 62.5 kg·m^2/s — Using formulas for moment of inertia, rotational kinetic energy and angular momentum for solid cylinder, values are calculated.

Question 12

12 (a) 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 child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Accepted Answer

(a) Given:
Initial angular speed, ω_i = 40 rev/min = (40 × 2π) / 60 = (4π/3) rad/s
Initial moment of inertia, I_i
Final moment of inertia, I_f = (2/5) I_i

Since no external torque, angular momentum conserved:
L_i = L_f
I_i ω_i = I_f ω_f

ω_f = (I_i / I_f) ω_i = (I_i / (2/5) I_i) ω_i = (5/2) ω_i = (5/2) × (4π/3) = (10π/3) rad/s ≈ 10.47 rad/s

(b) Initial kinetic energy:
K_i = (1/2) I_i ω_i^2

Final kinetic energy:
K_f = (1/2) I_f ω_f^2 = (1/2) × (2/5) I_i × ( (5/2) ω_i )^2 = (1/2) × (2/5) I_i × (25/4) ω_i^2 = (1/2) × (2/5) × (25/4) I_i ω_i^2 = (25/20) I_i ω_i^2 = (5/4) K_i

Thus, K_f = 1.25 K_i > K_i

The increase in kinetic energy is due to the work done by the child in pulling his arms inward, which reduces moment of inertia and increases angular speed, thereby increasing rotational kinetic energy. — Conservation of angular momentum relates initial and final angular speeds. The increase in kinetic energy comes from internal work done by the child.

Question 13

13 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 N ? What is the linear acceleration of the rope ? Assume that there is no slipping.

Accepted Answer

Given:
Mass, M = 3 kg
Radius, R = 0.40 m
Force, F = 30 N

Moment of inertia of hollow cylinder about its axis:
I = M R^2 = 3 × (0.4)^2 = 0.48 kg·m^2

Torque τ = F × R = 30 × 0.4 = 12 N·m

Angular acceleration:
α = τ / I = 12 / 0.48 = 25 rad/s^2

Linear acceleration of rope:
a = α R = 25 × 0.4 = 10 m/s^2 — Torque causes angular acceleration; linear acceleration of rope is related to angular acceleration by a = α R.

Question 14

14 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 velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

Accepted Answer

Given:
Angular speed, ω = 200 rad/s
Torque, τ = 180 N·m

Power P = τ × ω = 180 × 200 = 36000 W = 36 kW

Therefore, the power required by the engine is 36 kW. — Power in rotational motion is product of torque and angular velocity.

Question 15

15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Accepted Answer

Let the mass of the original disk be M.

Mass of hole = m = (area of hole / area of disk) × M = (π (R/2)^2) / (π R^2) × M = (1/4) M

Let the centre of the original disk be at origin O.

Position vector of hole's centre from O is r = R/2 along x-axis.

The centre of mass of the remaining body is given by:

R_cm = (M × 0 - m × r) / (M - m) = (0 - (1/4) M × (R/2)) / (M - (1/4) M) = (- (1/4) M × (R/2)) / ( (3/4) M ) = - (1/4) × (R/2) / (3/4) = - (1/4) × (R/2) × (4/3) = - R/6

The negative sign indicates the centre of gravity shifts opposite to the hole's centre.

Therefore, the centre of gravity of the remaining body is at a distance R/6 from the original centre, opposite to the hole's centre. — Using the principle of superposition, the hole is treated as negative mass at its centre, and the resultant centre of mass is calculated accordingly.

Systems Of Particles And Rotational Motion

Systems Of Particles And Rotational Motion — Study Notes

Introduction

Centre of Mass

Linear Momentum of a System of Particles

Practice Questions — Systems Of Particles And Rotational Motion

All 7 Chapters in Physics Part-I