JEE Main Rotational Motion PYQ

11. (JEE Main 2023 (Online) 31st January Morning Shift )

A solid sphere of mass $1 kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{- 3} J$ . The speed of the centre of mass of the sphere is __________ ${cm s}^{- 1}$

Correct answer is (10)

$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = 7 \times 10^{- 3}$

$\Rightarrow$ $\frac{1}{2} {mv}^{2} + \frac{1}{2} (\frac{2}{5} {MR}^{2}) {(\frac{V}{R})}^{2} = 7 \times 10^{- 3}$

$\Rightarrow$ $\frac{1}{2} {MV}^{2} [1 + \frac{2}{5}] = 7 \times 10^{- 3}$

$\Rightarrow$ $\frac{1}{2} (1) (V^{2}) (\frac{7}{5}) = 7 \times 10^{- 3}$

$\Rightarrow$ $V^{2} = 10^{- 2}$

$\Rightarrow$ $V = 10^{- 1} = 0.1 m / s = 10 cm / s$

12. (JEE Main 2023 (Online) 30th January Evening Shift )

A uniform disc of mass $0.5 kg$ and radius $r$ is projected with velocity $18 m / s$ at $t = 0$ s on a rough horizontal surface. It starts off with a purely sliding motion at $t = 0 s$ . After $2 s$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2 s$ will be __________ $J$ (given, coefficient of friction is $0.3$ and $g = 10 m / s^{2}$ ).

JEE Main 2023 (Online) 30th January Evening Shift Physics - Rotational Motion Question 21 English

Correct answer is (54)

$v = v_{0} - μ g t$

$\Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12$ m/s

$\Rightarrow$ Kinetic energy $= \frac{1}{2} m v^{2} + \frac{1}{2} \frac{m v^{2}}{2}$

$= \frac{3}{4} m v^{2} = \frac{3}{4} \times 0.5 \times 144 J = 54 J$

13. (JEE Main 2023 (Online) 29th January Morning Shift )

A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be _______ ms $^{- 1}$ .

Correct answer is (40)

$\frac{1}{2} m v_{cm}^{2} + \frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v_{cm}^{2}}{R^{2}} = 2240 J$

$\begin{aligned} \frac{7}{10} m v_{cm}^{2} = 2240 \\ v_{cm} = \sqrt{\frac{2240 \times 10}{7 \times 2}} = 40 m / \sec \end{aligned}$

14. (JEE Main 2022 (Online) 25th July Morning Shift)

A solid cylinder and a solid sphere, having same mass $M$ and radius $R$ , roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be :

A. $\sqrt{\frac{5}{3}}$

B. $\sqrt{\frac{4}{5}}$

C. $\sqrt{\frac{3}{5}}$

D. $\sqrt{\frac{14}{15}}$

Correct Answer is Option (D)

$a = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}}$

$v = \sqrt{\frac{2 S g \sin θ}{1 + \frac{K^{2}}{R^{2}}}}$

$\Rightarrow \frac{v_{c}}{v_{s s}} \sqrt{\frac{1 + \frac{K_{s s}^{2}}{R^{2}}}{1 + \frac{K_{c}^{2}}{R^{2}}}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + \frac{1}{2}}}$

$\Rightarrow \sqrt{\frac{\frac{7}{5}}{\frac{3}{2}}} = \sqrt{\frac{14}{15}}$

15. (JEE Main 2022 (Online) 26th June Evening Shift)

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

A. $\frac{2}{5}$

B. $\frac{2}{7}$

C. $\frac{1}{5}$

D. $\frac{7}{10}$

Correct Answer is Option (B)

$K E_{R} = \frac{1}{2} l w^{2}$

$= \frac{1}{2} \times \frac{2}{5} \times ω^{2} \times (m R^{2})$

$K E_{t o t a l} = \frac{1}{2} \times \frac{7}{5} \times m R^{2} \times ω^{2}$

$∴$ $\frac{K E_{R}}{K E_{t o t a l}} = \frac{2}{7}$