Work Required to Stop Rotating Objects – Rankers Physics

Work Required to Stop Rotating Objects

Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation (2018)

$W_B > W_A > W_C$
$W_A > W_B > W_C$
$W_C > W_B > W_A$
$W_A > W_C > W_B$

Solution:

Work required equals rotational kinetic energy $\frac{1}{2}I\omega^2$. Comparing moments of inertia, $I_C = MR^2 > I_B = 0.5MR^2 > I_A = 0.4MR^2$, leading to $W_C > W_B > W_A$.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Questions

Explore conceptually related problems

1. The instantaneous angular position of a point on a rotating wheel is given by the equation $$\theta (t) = 2t^{3} - 6t^{2}$$. The torque on the wheel becomes zero at: (2011 Pre)
2. A particle of mass \(m\) is moving on a circle of radius \(R\) with kinetic energy \(K\). Then angular momentum of particle about centre of circle will be:
3. Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$. (2020)
4. From a circular ring of mass \(M\) and radius \(R\) an arc corresponding to a \(90^\circ\) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is \(K\) times \(MR^2\). Then the value of \(K\) is
5. The moment of inertia of a body about a given axis is $1.2 \text{ kgm}^2$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500 \text{ joule}$, an angular acceleration of $25 \text{ rad/sec}^2$ must be applied about that axis for a duration of: (1990)