DPP - Work, Power & Energy in Rotational Motion
1. A body having moment of inertia about its axis of rotation equal to 3 kg·m² is rotating with angular velocity equal to 3 rad/s. Kinetic energy of this rotating body is the same as that of body of mass 27 kg moving with a speed of
Correct Answer: (a) 1 m/s
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times 3 \times (3)^2 = 13.5\) J
Translational KE = \(\frac{1}{2} m v^2 = \frac{1}{2} \times 27 \times v^2 = 13.5\) J
Solving: \(v^2 = 1\) → \(v = 1\) m/s
Translational KE = \(\frac{1}{2} m v^2 = \frac{1}{2} \times 27 \times v^2 = 13.5\) J
Solving: \(v^2 = 1\) → \(v = 1\) m/s
2. A solid sphere of mass 500 gm and radius 10 cm rolls without slipping with the velocity 20 cm/s. The total kinetic energy of the sphere will be
Correct Answer: (a) 0.014 J
For a solid sphere: \(I = \frac{2}{5} m R^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\)
Since \(\omega = \frac{v}{R}\), KE = \(\frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{7}{10} m v^2\)
m = 0.5 kg, v = 0.2 m/s
KE = \(\frac{7}{10} \times 0.5 \times (0.2)^2 = 0.014\) J
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\)
Since \(\omega = \frac{v}{R}\), KE = \(\frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{7}{10} m v^2\)
m = 0.5 kg, v = 0.2 m/s
KE = \(\frac{7}{10} \times 0.5 \times (0.2)^2 = 0.014\) J
3. A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with angular velocity of 20 rad/s. Its kinetic energy is
Correct Answer: (d) 250 J
For a ring rotating about its diameter: \(I = \frac{1}{2} m R^2\)
KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{1}{2} \times 10 \times (0.5)^2 \times (20)^2\)
= \(\frac{1}{4} \times 10 \times 0.25 \times 400 = 250\) J
KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{1}{2} \times 10 \times (0.5)^2 \times (20)^2\)
= \(\frac{1}{4} \times 10 \times 0.25 \times 400 = 250\) J
4. A thin hollow cylinder open at both ends: (i) Sliding without rotating (ii) Rolls without slipping, with the same speed. The ratio of kinetic energy in the two cases is
Correct Answer: (c) 1 : 2
Case (i): Only translational KE = \(\frac{1}{2} m v^2\)
Case (ii): For hollow cylinder, \(I = m R^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} m R^2 \times \frac{v^2}{R^2} = m v^2\)
Ratio = \(\frac{\frac{1}{2} m v^2}{m v^2} = \frac{1}{2}\)
Case (ii): For hollow cylinder, \(I = m R^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} m R^2 \times \frac{v^2}{R^2} = m v^2\)
Ratio = \(\frac{\frac{1}{2} m v^2}{m v^2} = \frac{1}{2}\)
5. Ratio of translatory kinetic energy and rotational kinetic energy in the motion of a disc is
Correct Answer: (c) 1 : 2
For a disc: \(I = \frac{1}{2} m R^2\)
Translational KE = \(\frac{1}{2} m v^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{1}{2} m R^2 \times \frac{v^2}{R^2} = \frac{1}{4} m v^2\)
Ratio = \(\frac{\frac{1}{2} m v^2}{\frac{1}{4} m v^2} = 2 : 1\) or 1 : 2 depending on order
Translational KE = \(\frac{1}{2} m v^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{1}{2} m R^2 \times \frac{v^2}{R^2} = \frac{1}{4} m v^2\)
Ratio = \(\frac{\frac{1}{2} m v^2}{\frac{1}{4} m v^2} = 2 : 1\) or 1 : 2 depending on order
6. The ratio of rotational and translatory kinetic energies of a sphere is
Correct Answer: (d) 2/7
For a solid sphere: \(I = \frac{2}{5} m R^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Translational KE = \(\frac{1}{2} m v^2\)
Ratio = \(\frac{\frac{1}{5} m v^2}{\frac{1}{2} m v^2} = \frac{2}{5}\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Translational KE = \(\frac{1}{2} m v^2\)
Ratio = \(\frac{\frac{1}{5} m v^2}{\frac{1}{2} m v^2} = \frac{2}{5}\)
7. A body of moment of inertia of 3 kg·m² rotating with an angular velocity of 2 rad/sec has the same kinetic energy as a mass of 12 kg moving with a velocity of
Correct Answer: (d) 1 m/s
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times 3 \times (2)^2 = 6\) J
Translational KE = \(\frac{1}{2} m v^2 = \frac{1}{2} \times 12 \times v^2 = 6\) J
Solving: \(v^2 = 1\) → \(v = 1\) m/s
Translational KE = \(\frac{1}{2} m v^2 = \frac{1}{2} \times 12 \times v^2 = 6\) J
Solving: \(v^2 = 1\) → \(v = 1\) m/s
8. A solid spherical ball rolls on an inclined plane without slipping. The ratio of rotational energy and total energy is
Correct Answer: (c) 2/7
For a solid sphere: \(I = \frac{2}{5} m R^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Total KE = Translational KE + Rotational KE = \(\frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2\)
Ratio = \(\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2} = \frac{2}{7}\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Total KE = Translational KE + Rotational KE = \(\frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2\)
Ratio = \(\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2} = \frac{2}{7}\)
9. When a sphere of moment of inertia I about its centre of gravity and mass 'm' rolls from rest down an inclined plane without slipping, its kinetic energy is
Correct Answer: (c) \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\)
For a rolling object without slipping, the total kinetic energy is the sum of translational kinetic energy and rotational kinetic energy.
10. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be
Correct Answer: (b) \(\frac{K^2}{K^2 + R^2}\)
Moment of inertia \(I = m K^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} m K^2 \omega^2\)
Since \(\omega = \frac{v}{R}\), Rotational KE = \(\frac{1}{2} m K^2 \frac{v^2}{R^2}\)
Translational KE = \(\frac{1}{2} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} m K^2 \frac{v^2}{R^2} = \frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})\)
Fraction = \(\frac{\frac{1}{2} m K^2 \frac{v^2}{R^2}}{\frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})} = \frac{K^2}{R^2 + K^2}\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} m K^2 \omega^2\)
Since \(\omega = \frac{v}{R}\), Rotational KE = \(\frac{1}{2} m K^2 \frac{v^2}{R^2}\)
Translational KE = \(\frac{1}{2} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} m K^2 \frac{v^2}{R^2} = \frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})\)
Fraction = \(\frac{\frac{1}{2} m K^2 \frac{v^2}{R^2}}{\frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})} = \frac{K^2}{R^2 + K^2}\)
11. A metre stick of mass 400 gm is pivoted at one end and displaced through an angle 60°. The increase in its potential energy is
Correct Answer: (d) 1 J
For a rod pivoted at one end, the center of mass is at L/2 from the pivot.
Height change of center of mass = \(\frac{L}{2} (1 - \cos\theta)\)
L = 1 m, m = 0.4 kg, θ = 60°
Δh = \(\frac{1}{2} (1 - \cos 60°) = \frac{1}{2} (1 - 0.5) = 0.25\) m
ΔPE = m g Δh = 0.4 × 10 × 0.25 = 1 J
Height change of center of mass = \(\frac{L}{2} (1 - \cos\theta)\)
L = 1 m, m = 0.4 kg, θ = 60°
Δh = \(\frac{1}{2} (1 - \cos 60°) = \frac{1}{2} (1 - 0.5) = 0.25\) m
ΔPE = m g Δh = 0.4 × 10 × 0.25 = 1 J
12. If 'I' is the moment of inertia of a body and 'ω' is its angular velocity, then its rotational kinetic energy is
Correct Answer: (c) \(\frac{1}{2} I \omega^2\)
This is the standard formula for rotational kinetic energy, analogous to \(\frac{1}{2} m v^2\) for translational kinetic energy.
13. The M.I. of a body about the given axis is 1.2 kg·m² initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad/sec² must be applied about that axis for duration of
Correct Answer: (b) 2 sec
Rotational KE = \(\frac{1}{2} I \omega^2 = 1500\) J
\(\frac{1}{2} \times 1.2 \times \omega^2 = 1500\) → \(\omega^2 = 2500\) → \(\omega = 50\) rad/s
Using \(\omega = \alpha t\) → \(50 = 25 \times t\) → \(t = 2\) s
\(\frac{1}{2} \times 1.2 \times \omega^2 = 1500\) → \(\omega^2 = 2500\) → \(\omega = 50\) rad/s
Using \(\omega = \alpha t\) → \(50 = 25 \times t\) → \(t = 2\) s
14. The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with a velocity of 2 m/s without slipping is 32.8 joule. The radius of gyration of the body is
Correct Answer: (d) 0.4 m
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2\)
Since \(\omega = \frac{v}{R}\) and \(I = m K^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} m K^2 \frac{v^2}{R^2} = \frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})\)
\(32.8 = \frac{1}{2} \times 10 \times (2)^2 \times (1 + \frac{K^2}{(0.5)^2})\)
\(32.8 = 20 \times (1 + 4K^2)\) → \(1 + 4K^2 = 1.64\) → \(4K^2 = 0.64\) → \(K^2 = 0.16\) → \(K = 0.4\) m
Since \(\omega = \frac{v}{R}\) and \(I = m K^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} m K^2 \frac{v^2}{R^2} = \frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})\)
\(32.8 = \frac{1}{2} \times 10 \times (2)^2 \times (1 + \frac{K^2}{(0.5)^2})\)
\(32.8 = 20 \times (1 + 4K^2)\) → \(1 + 4K^2 = 1.64\) → \(4K^2 = 0.64\) → \(K^2 = 0.16\) → \(K = 0.4\) m
15. A hollow sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity v = 0.5 m/s. The kinetic energy of the sphere is
Correct Answer: (d) 0.42 J
For a hollow sphere: \(I = \frac{2}{3} m R^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{3} m R^2 \times \frac{v^2}{R^2} = \frac{5}{6} m v^2\)
m = 2 kg, v = 0.5 m/s
KE = \(\frac{5}{6} \times 2 \times (0.5)^2 = 0.4167 \approx 0.42\) J
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{3} m R^2 \times \frac{v^2}{R^2} = \frac{5}{6} m v^2\)
m = 2 kg, v = 0.5 m/s
KE = \(\frac{5}{6} \times 2 \times (0.5)^2 = 0.4167 \approx 0.42\) J
16. The total kinetic energy of rolling solid sphere having translational velocity v is
Correct Answer: (a) \(\frac{7}{10} m v^2\)
For a solid sphere: \(I = \frac{2}{5} m R^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{7}{10} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{7}{10} m v^2\)
17. A spherical solid ball of 1 kg mass and radius 3 cm is rotating about an axis passing through its centre with an angular velocity of 50 radian/s. The kinetic energy of rotation is
Correct Answer: (b) 90 J
For a solid sphere: \(I = \frac{2}{5} m R^2 = \frac{2}{5} \times 1 \times (0.03)^2 = 0.00036\) kg·m²
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.00036 \times (50)^2 = 0.45\) J
Note: There might be an error in the provided options. The calculation gives 0.45 J, not 90 J.
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.00036 \times (50)^2 = 0.45\) J
Note: There might be an error in the provided options. The calculation gives 0.45 J, not 90 J.
18. A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is
Correct Answer: (c) 2/7
For a solid sphere: \(I = \frac{2}{5} m R^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2\)
Fraction = \(\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2} = \frac{2}{7}\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2\)
Fraction = \(\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2} = \frac{2}{7}\)
19. A flywheel is in the form of solid circular wheel of mass 72 kg and radius of 0.5m and it takes 70 r.p.m., then the energy of revolution is
Correct Answer: (b) 240 J
For a solid disc: \(I = \frac{1}{2} m R^2 = \frac{1}{2} \times 72 \times (0.5)^2 = 9\) kg·m²
Angular velocity ω = \(2\pi \times \frac{70}{60} = \frac{7\pi}{3}\) rad/s
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times 9 \times (\frac{7\pi}{3})^2 \approx 240\) J
Angular velocity ω = \(2\pi \times \frac{70}{60} = \frac{7\pi}{3}\) rad/s
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times 9 \times (\frac{7\pi}{3})^2 \approx 240\) J
20. A disc of moment of inertia \(I = 5\) kg·m² is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done
Correct Answer: (a) 1467 J
ω₁ = \(2\pi \times \frac{600}{60} = 20\pi\) rad/s
ω₂ = \(2\pi \times \frac{300}{60} = 10\pi\) rad/s
Work done = Change in KE = \(\frac{1}{2} I (\omega_1^2 - \omega_2^2) = \frac{1}{2} \times 5 \times ((20\pi)^2 - (10\pi)^2)\)
= \(\frac{5}{2} \times (400\pi^2 - 100\pi^2) = \frac{5}{2} \times 300\pi^2 = 750\pi^2 \approx 1467\) J
ω₂ = \(2\pi \times \frac{300}{60} = 10\pi\) rad/s
Work done = Change in KE = \(\frac{1}{2} I (\omega_1^2 - \omega_2^2) = \frac{1}{2} \times 5 \times ((20\pi)^2 - (10\pi)^2)\)
= \(\frac{5}{2} \times (400\pi^2 - 100\pi^2) = \frac{5}{2} \times 300\pi^2 = 750\pi^2 \approx 1467\) J
21. A circular disc of mass 0.41 kg and radius 10 m rolls without slipping with a velocity of 2 m/s. The total kinetic energy of disc is
Correct Answer: (b) 1.23 J
For a disc: \(I = \frac{1}{2} m R^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{1}{2} m R^2 \times \frac{v^2}{R^2} = \frac{3}{4} m v^2\)
m = 0.41 kg, v = 2 m/s
KE = \(\frac{3}{4} \times 0.41 \times (2)^2 = 1.23\) J
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{1}{2} m R^2 \times \frac{v^2}{R^2} = \frac{3}{4} m v^2\)
m = 0.41 kg, v = 2 m/s
KE = \(\frac{3}{4} \times 0.41 \times (2)^2 = 1.23\) J
22. A sphere of mass 0.5 kg and diameter 1m rolls without sliding with a constant velocity of 5 m/s, calculate what is the ratio of the rotational K.E. to the total kinetic energy of the sphere
Correct Answer: (d) 2/7
For a solid sphere: \(I = \frac{2}{5} m R^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2\)
Ratio = \(\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2} = \frac{2}{7}\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2} = \frac{1}{5} m v^2\)
Total KE = \(\frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2\)
Ratio = \(\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2} = \frac{2}{7}\)
23. A solid homogeneous sphere is moving on a rough horizontal surface partly rolling and partly sliding. During this kind of motion of the sphere
Correct Answer: (b) The angular momentum of the sphere about the point of contact with the plane is conserved
When a sphere is partly rolling and partly sliding, friction acts at the point of contact. The torque of this friction about the point of contact is zero, so angular momentum about that point is conserved.
24. The moments of inertia of two freely rotating bodies A and B are I_A and I_B respectively. I_A > I_B and their angular momenta are equal. If K_A and K_B are their kinetic energies, then
Correct Answer: (c) K_A < K_B
Rotational KE = \(\frac{L^2}{2I}\) where L is angular momentum
Since L is same for both, KE ∝ \(\frac{1}{I}\)
I_A > I_B ⇒ K_A < K_B
Since L is same for both, KE ∝ \(\frac{1}{I}\)
I_A > I_B ⇒ K_A < K_B
25. A ring of radius r and mass m rotates about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Its kinetic energy is
Correct Answer: (c) \(\frac{1}{2} m r^2 \omega^2\)
For a ring about an axis through its center perpendicular to its plane: \(I = m r^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} m r^2 \omega^2\)
Rotational KE = \(\frac{1}{2} I \omega^2 = \frac{1}{2} m r^2 \omega^2\)