A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_{i}$. Another disk of moment of inertia $I_{b}$ is dropped coaxially into the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_{f}$. The energy lost by the initially rotating disc due to friction is:
(2010)
1. $$\frac{1}{2}\frac{I_{b}^{2}}{(I_{t}+I_{b})}\omega_{i}^{2}$$
2. $$\frac{1}{2}\frac{I_{t}^{2}}{(I_{t}+I_{b})}\omega_{i}^{2}$$
3. $$\frac{I_{b}-I_{t}}{(I_{t}+I_{b})}\omega_{i}^{2}$$
4. $$\frac{1}{2}\frac{I_{b}I_{t}}{(I_{t}+I_{b})}\omega_{i}^{2}$$
View Answer
By conservation of angular momentum, $I_{t}\omega_{i} = (I_{t}+I_{b})\omega_{f}$, yielding $\omega_{f} = \frac{I_{t}\omega_{i}}{I_{t}+I_{b}}$. The loss in kinetic energy is $\Delta K = \frac{1}{2}I_{t}\omega_{i}^{2} - \frac{1}{2}(I_{t}+I_{b})\omega_{f}^{2}$. Substituting $\omega_{f}$ simplifies to $\Delta K = \frac{1}{2}\frac{I_{t}I_{b}}{(I_{t}+I_{b})}\omega_{i}^{2}$.
A wheel having moment of inertia $2 \text{ kg-m}^2$ about its vertical axis, rotates at the rate of $60 \text{ rpm}$ about the axis. The torque which can stop the wheel’s rotation in one minute would be:
(2004)
1. $\frac{\pi}{12} \text{ N-m}$
2. $\frac{\pi}{15} \text{ N-m}$
3. $\frac{\pi}{18} \text{ N-m}$
4. $\frac{2\pi}{15} \text{ N-m}$
View Answer
Initial angular velocity $\omega_0 = 60 \text{ rpm} = \frac{60 \times 2\pi}{60} = 2\pi \text{ rad/s}$. Final $\omega = 0$. Time $t = 60 \text{ s}$.
Angular acceleration $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 2\pi}{60} = -\frac{\pi}{30} \text{ rad/s}^2$.
Required torque $\tau = I|\alpha| = 2 \times \frac{\pi}{30} = \frac{\pi}{15} \text{ N-m}$.
A thin circular ring of mass $M$ and radius ‘$r$’ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:
(2003)
1. $\frac{M\omega}{4m}$
2. $\frac{M\omega}{M + 4m}$
3. $\frac{(M + 4m)\omega}{M}$
4. $\frac{(M + 4m)\omega}{M + 4m}$
View Answer
By conservation of angular momentum, $I_{initial}\omega_{initial} = I_{final}\omega_{final}$.
Initial moment of inertia $I_i = Mr^2$. Final moment of inertia $I_f = Mr^2 + 4mr^2$.
Thus, $$Mr^2 \omega = (M + 4m)r^2 \omega' \implies \omega' = \frac{M\omega}{M + 4m}$$.
A circular ring of mass $M$ and radius $R$ is rotating about its axis with constant angular velocity $\omega$. Two particles each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The angular velocity of the ring will now become:
(1998)
1. $\frac{M - 2m}{m\omega}$
2. $\frac{m\omega}{M - 2m}$
3. $\frac{M\omega}{M + 2m}$
4. $\frac{M - 2m}{m}$
View Answer
Since the particles are attached gently, external torque is zero, meaning angular momentum is conserved.
$I_1\omega_1 = I_2\omega_2 \implies (MR^2)\omega = (MR^2 + 2mR^2)\omega'$.
Solving for the new angular velocity yields $\omega' = \frac{M\omega}{M + 2m}$.
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:
(2017-Delhi)
1. $\frac{1}{4}I(\omega_1-\omega_2)^2$
2. $I(\omega_1-\omega_2)^2$
3. $\frac{1}{8}I(\omega_1-\omega_2)^2$
4. $\frac{1}{2}I(\omega_1-\omega_2)^2$
View Answer
By conservation of angular momentum, the final common angular velocity is $\omega = \frac{\omega_1 + \omega_2}{2}$. The loss in rotational kinetic energy is $Delta E = E_i - E_f = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2 - 2 \cdot \left(\frac{1}{2}I\omega^2\right)$, which simplifies to $\frac{1}{8}I(\omega_1-\omega_2)^2$.
A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
(2016 – I)
1. Disk
2. Sphere
3. Both reach at the same time
4. Depends on their masses
View Answer
Acceleration of a rolling body on an inclined plane is given by $$a = \frac{g sin\theta}{1 + I/(mR^2)}$$. Since the acceleration is independent of mass and depends only on the geometry (moment of inertia factor), the sphere has a smaller inertia factor than the disk, meaning the sphere has a greater acceleration and reaches the bottom first.
Small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is:
(2013)
1. Disc
2. Ring
3. Solid sphere
4. Hollow sphere
View Answer
Using energy conservation, initial kinetic energy equals potential energy at max height: $\frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) = mgH$. Substituting $H = \frac{3v^2}{4g}$, we get $1 + \frac{I}{mR^2} = 2$, which gives $\frac{I}{mR^2} = 1$. This corresponds to a ring.