System of Particles and Rotational Motion — Practice

Question 1

Masses 2 kg and 3 kg are at (1,0) and (4,0) m. Find the centre of mass.

Accepted Answer

X_cm = (2 x 1 + 3 x 4)/(2 + 3) = 14/5 = 2.8 m from the origin.

Question 2

A solid cylinder of mass 20 kg and radius 0.25 m rotates about its own axis. Find its moment of inertia and its radius of gyration.

Accepted Answer

For a solid cylinder about its axis, I = MR^2/2 = 20 x 0.25^2 / 2 = 0.625 kg m^2. Radius of gyration k = sqrt(I/M) = sqrt(0.625/20) = sqrt(0.03125) = 0.177 m. Note k is smaller than R, since the mass is distributed at all distances from the axis, not concentrated at the rim.

Question 3

A torque of 10 N m acts on a body with I = 5 kg m^2 for 4 s from rest. Find the angular velocity attained.

Accepted Answer

alpha = tau/I = 10/5 = 2 rad/s^2. omega = alpha t = 2 x 4 = 8 rad/s.

Question 4

A diver with I = 15 kg m^2 spins at 3 rad/s tucked, then straightens to I = 25 kg m^2. Find the new angular velocity.

Accepted Answer

Conservation of angular momentum: I1 omega1 = I2 omega2. 15 x 3 = 25 x omega2, so omega2 = 1.8 rad/s.

Question 5

A solid sphere rolls down a 10 m high incline. Find its speed at the bottom (g = 10 m/s^2).

Accepted Answer

Mgh = (1/2)MV^2 + (1/2)I omega^2 with I = (2/5)MR^2 and omega = V/R, giving Mgh = (7/10)MV^2. V = sqrt(10gh/7) = sqrt(1000/7) = 11.95 m/s.

Question 6

A uniform rod of length 2 m and weight 20 N rests horizontally on two supports, one at each end. A 30 N weight hangs 0.5 m from the left end. Find the reaction force at each support.

Accepted Answer

Taking moments about the left support: the rod's own weight (20 N) acts at its centre, 1 m away; the 30 N weight acts at 0.5 m. R_right x 2 = (20 x 1) + (30 x 0.5) = 20 + 15 = 35, so R_right = 17.5 N. Total weight is 50 N, so R_left = 50 - 17.5 = 32.5 N.

Question 7

A man stands at the centre of a rotating platform holding a 5 kg mass in each hand, 0.8 m from the axis. The platform-plus-man system (excluding the masses) has I = 2 kg m^2 and spins at 2 rad/s. He pulls both masses in to 0.2 m from the axis. Find the new angular speed.

Accepted Answer

I_initial = 2 + 2(5)(0.8)^2 = 2 + 6.4 = 8.4 kg m^2. I_final = 2 + 2(5)(0.2)^2 = 2 + 0.4 = 2.4 kg m^2. Conservation of angular momentum: 8.4 x 2 = 2.4 x omega2, so omega2 = 16.8/2.4 = 7 rad/s.

Question 8

Three point masses 1 kg, 2 kg and 3 kg are placed at (0,0), (2,0) and (0,3) m respectively. Find the coordinates of the centre of mass.

Accepted Answer

X_cm = (1x0 + 2x2 + 3x0)/6 = 4/6 = 0.67 m. Y_cm = (1x0 + 2x0 + 3x3)/6 = 9/6 = 1.5 m. Centre of mass is at (0.67, 1.5) m.

Question 9

Two point masses of 2 kg each are fixed at the ends of a light rod of length 1 m. Find the moment of inertia of the system about an axis through the centre of the rod, perpendicular to it.

Accepted Answer

Each mass is 0.5 m from the axis. I = 2 x (2 x 0.5^2) = 2 x 2 x 0.25 = 1 kg m^2. Being a light rod, only the two point masses contribute.

Practice — System of Particles and Rotational Motion

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