A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. The objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by:
(2010 Mains)
1. $\frac{(M+2m)\omega}{2m}$
2. $\frac{2M\omega}{M+2m}$
3. $\frac{(M+2m)\omega}{M}$
4. $\frac{M\omega}{M+2m}$
View Answer
By conservation of angular momentum, $I_{1}\omega_{1} = I_{2}\omega_{2}$. Initially, $I_{1} = Mr^{2}$. Finally, the moment of inertia is $I_{2} = Mr^{2} + 2mr^{2} = (M+2m)r^{2}$. Equating the two yields $Mr^{2}\omega = (M+2m)r^{2}\omega_{2}$, so $\omega_{2} = \frac{M\omega}{M+2m}$.
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 particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt{2}$ in the XOY plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is:
(1991)
1. $60 \text{ units}$
2. $40\sqrt{2} \text{ units}$
3. Zero
4. $7.5 \text{ units}$
View Answer
Angular momentum $L = mvr_{\perp}$. The line equation is $X - Y + 4 = 0$.
The perpendicular distance $r_{\perp}$ from the origin $(0,0)$ to the line is $\frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Thus, $L = 5 \times (3\sqrt{2}) \times (2\sqrt{2}) = 60 \text{ units}$.
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 disc is rotating with angular speed $\omega$. If a child sits on it, what is conserved:
(2002)
1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. Potential energy
View Answer
When the child sits on the rotating disc gently, no external torque acts on the system.
According to Newton's second law for rotation, if net external torque is zero, the total angular momentum of the system remains conserved.
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 $\omega' = \frac{M\omega}{M + 2m}$.
If a ladder is not in balance against a smooth vertical wall, then it can be made in balance by:
(1998)
1. Decreasing the length of ladder
2. Increasing the length of ladder
3. Increasing the angle of inclination
4. Decreasing the angle of inclination
View Answer
For equilibrium, the required frictional force at the base is $f = \frac{mg}{2} \cot\theta$, where $\theta$ is the angle of inclination with the horizontal.
To prevent slipping, $f$ must be less than or equal to the limiting friction $\mu mg$.
To decrease the required friction $f$, we must decrease $\cot\theta$, which means increasing the angle of inclination $\theta$.
A couple produces:
(1997)
1. Linear and rotational motion
2. No motion
3. Purely linear motion
4. Purely rotational motion
View Answer
A couple consists of two equal and opposite parallel forces whose lines of action do not coincide.
The net force is zero, so there is no translational (linear) acceleration.
However, there is a net torque, which produces purely rotational motion.
Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$
(1997)
1. $-21\hat{i} + 4\hat{j} + 4\hat{k}$
2. $-14\hat{i} + 34\hat{j} - 16\hat{k}$
3. $14\hat{i} - 38\hat{j} + 16\hat{k}$
4. $4\hat{i} + 4\hat{j} + 6\hat{k}$
View Answer
Torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Using the determinant method: $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$.
$= \hat{i}(15 - 1) - \hat{j}(35 - (-3)) + \hat{k}(7 - (-9)) = 14\hat{i} - 38\hat{j} + 16\hat{k}$.