A force $\vec{F} = \alpha\hat{i} + 3\hat{j} + 9\hat{k}$ is acting at a point $\vec{r} = 2\hat{i} – 6\hat{j} – 12\hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: (2015 Re)
1. $1$
2. $-1$
3. $2$
4. Zero
View Answer
For angular momentum to be conserved, torque $\vec{\tau} = \vec{r} \times \vec{F}$ must be zero, meaning $\vec{r}$ and $\vec{F}$ are collinear. Taking the ratio of their components: $\frac{2}{\alpha} = \frac{-6}{3} = \frac{-12}{9}$, which simplifies to $\frac{2}{\alpha} = -2$, giving $\alpha = -1$.
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)
1. $4 \text{ s}$
2. $2 \text{ s}$
3. $8 \text{ s}$
4. $10 \text{ s}$
View Answer
Using $K = \frac{1}{2} I \omega^2$, we substitute the values to get $1500 = \frac{1}{2}(1.2) \omega^2$, giving $\omega = 50 \text{ rad/s}$. Applying kinematics equation $\omega = \omega_0 + \alpha t$, we find $50 = 0 + 25t$, yielding $t = 2 \text{ s}$.
A ring of mass $m$ and radius $r$ rotates about an axis passing through its centre and perpendicular to its plane with angular velocity $\omega$. Its kinetic energy is:
(1988)
1. $\frac{1}{2} mr^2 \omega^2$
2. $mr \omega^2$
3. $mr^2 \omega^2$
4. $\frac{1}{2} mr \omega^2$
View Answer
For a ring rotating about its central perpendicular axis, the moment of inertia is $I = mr^2$. The rotational kinetic energy is defined as $K = \frac{1}{2} I \omega^2$. Substituting $I$, we get $K = \frac{1}{2} mr^2 \omega^2$.
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)
1. $6 \hat{j} \text{ N m}$
2. $-6 \hat{i} \text{ N m}$
3. $6 \hat{k} \text{ N m}$
4. $6 \hat{i} \text{ N m}$
View Answer
Torque is given by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Substituting the given vectors, $\vec{\tau} = (2\hat{k}) \times (3\hat{j}) = 6(\hat{k} \times \hat{j})$. Since $\hat{k} \times \hat{j} = -\hat{i}$, the torque is $-6\hat{i} \text{ N m}$.
A solid cylinder of mass $2 \text{ kg}$ and radius $4 \text{ cm}$ is rotating about its axis at the rate of $3 \text{ rpm}$. The torque required to stop after $2\pi$ revolutions is
(2019)
1. $2 \times 10^{-6} \text{ N m}$
2. $2 \times 10^{-3} \text{ N m}$
3. $12 \times 10^{-4} \text{ N m}$
4. $2 \times 10^{6} \text{ N m}$
View Answer
Here $I = \frac{1}{2}MR^2 = 1.6 \times 10^{-3} \text{ kg m}^2$, $\omega_0 = 3 \times \frac{2\pi}{60} = \frac{\pi}{10} \text{ rad/s}$, and $\theta = 4\pi^2 \text{ rad}$. Using $\omega^2 = \omega_0^2 + 2\alpha\theta$, $\alpha = -\frac{1}{800} \text{ rad/s}^2$. The required torque magnitude is $\tau = I\alpha = 2 \times 10^{-6} \text{ N m}$.
A circular platform is mounted on a frictionless vertical axle. Its radius $R = 2\text{ m}$ and its moment of inertia about the axle is $200\text{ kg m}^{2}$. It is initially at rest. A $50\text{ kg}$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1\text{ ms}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is:
(2012 Mains)
1. $\pi\text{ s}$
2. $3\pi/2\text{ s}$
3. $2\pi\text{ s}$
4. $\pi/2\text{ s}$
View Answer
Angular velocity of man $\omega_{m} = \frac{v}{R} = 0.5\text{ rad/s}$. By conservation of angular momentum, $I_{p}\omega_{p} + I_{m}\omega_{m} = 0$, giving $\omega_{p} = -0.5\text{ rad/s}$. Relative angular velocity $\omega_{rel} = \omega_{m} - \omega_{p} = 1\text{ rad/s}$. Time $T = \frac{2\pi}{\omega_{rel}} = 2\pi\text{ s}$.
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along
(2012 Pre)
1. A line perpendicular to the plane of rotation
2. The line making an angle of $45^{\circ}$ to the plane of rotation
3. The radius
4. The tangent to the orbit
View Answer
Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$. According to the properties of the cross product, the vector $\vec{L}$ is directed perpendicular to the plane containing the position vector $\vec{r}$ and momentum vector $\vec{p}$.
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 $\omega_{2} = \frac{M\omega}{M+2m}$.