What is torque of the force $\vec{F} = 2\hat{i} – 3\hat{j} + 4\hat{k}$ acting at the point $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ about origin?
(1995)
1. $-6\hat{i} + 6\hat{j} - 12\hat{k}$
2. $-17\hat{i} + 6\hat{j} + 13\hat{k}$
3. $6\hat{i} - 6\hat{j} + 12\hat{k}$
4. $17\hat{i} - 6\hat{j} - 13\hat{k}$
View Answer
Torque is $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$.
$= \hat{i}(8 - (-9)) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) = 17\hat{i} - 6\hat{j} - 13\hat{k}$.
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.
A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
(2010 Mains)
1. Both together only when angle of inclination of plane is $45^{\circ}$
2. Both together
3. Hollow cylinder
4. Solid cylinder
View Answer
The acceleration of a rolling body depends on its moment of inertia ratio $I/mR^2$. The solid cylinder has a smaller moment of inertia ratio ($1/2$) compared to the hollow cylinder ($1$), giving it a higher acceleration and causing it to reach the bottom first.
A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle $theta$. The frictional force:
(2005)
1. Converts translational energy to rotational energy
2. Dissipates energy as heat
3. Decreases the rotational motion
4. Decreases the rotational and translational motion
View Answer
Static friction provides the necessary torque for rolling without slipping, converting translational kinetic energy into rotational kinetic energy without dissipating mechanical energy.