A disc of mass 1,kg and radius R is free to r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $m$ be the mass of the disc and the particle. The moment of inertia of the disc about the axis is $I_{	ext{disc}} = \frac{1}{2} mR^2$. The mo

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $m$ be the mass of the disc and the particle. The moment of inertia of the disc about the axis is $I_{	ext{disc}} = \frac{1}{2} mR^2$. The moment of inertia of the particle at distance $R$ is $I_{	ext{particle}} = mR^2$. The total moment of inertia is $I = I_{	ext{disc}} + I_{	ext{particle}} = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$.Using the law of conservation of mechanical energy,the loss in potential energy of the particle equals the gain in rotational kinetic energy of the system.The particle moves from the highest point to the lowest point,a vertical distance of $2R$.Loss in potential energy $= mg(2R) = 2mgR$.Gain in rotational kinetic energy $= \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{3}{2} mR^2 ight) \omega^2 = \frac{3}{4} mR^2 \omega^2$.Equating the two: $2mgR = \frac{3}{4} mR^2 \omega^2$.Solving for $\omega^2$: $\omega^2 = \frac{8g}{3R}$.Given $\omega = 4 \sqrt{\frac{x}{3R}}$,so $\omega^2 = 16 \left( \frac{x}{3R} ight) = \frac{16x}{3R}$.Equating the expressions for $\omega^2$: $\frac{16x}{3R} = \frac{8g}{3R} \implies 16x = 8g$.Substituting $g = 10 	ext{ m s}^{-2}$: $16x = 8(10) = 80$.$x = \frac{80}{16} = 5$.

Similar Questions

The angular velocity of a body is $\vec{\omega} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and a torque $\vec{\tau} = \hat{i} + 2\hat{j} + 3\hat{k}$ acts on it. The rotational power will be .......... $W$.

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

$A$ thin rod of length $L$ and mass $M$ is held vertically with one end on the floor and is allowed to fall. Find the velocity of the other end when it hits the floor,assuming that the end on the floor does not slip.

Vedclass Test Series

Exam Paper Generator

Online Exam Module