A solid spherical ball rolls on a table. Ratio of its rotational kinetic energy to total kinetic energy is:
(1994)
For a solid sphere, $I = \frac{2}{5}MR^2$. Rotational kinetic energy is $\frac{1}{2}Iomega^2 = \frac{1}{5}MR^2\omega^2$ and total kinetic energy is $\frac{7}{10}MR^2\omega^2$. Ratio is $\frac{2}{7}$.
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is:
(2006, 2005)
Using the parallel axis theorem, $I = I_{cm} + Md^2$. Here $I_{cm} = \frac{1}{2}MR^2$ and $d = R$, so $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
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)
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 $t = 2 \text{ s}$.
Moment of inertia of a uniform circular disc about a diameter is $I$. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be:
(1990)
Given $I_{\text{diameter}} = \frac{MR^2}{4} = I$, which means $MR^2 = 4I$. Using the parallel axis theorem, the moment of inertia about a perpendicular axis on the rim is $I_{\text{rim}} = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2 = \frac{3}{2} (4I) = 6I$.
A fly wheel rotating about fixed axis has a kinetic energy of $360 \text{ joule}$ when its angular speed is $30 \text{ rad/sec}$. The moment of inertia of the wheel about the axis of rotation is:
(1990)
The formula for rotational kinetic energy is $K = \frac{1}{2} I \omega^2$. Substituting the given values, $360 = \frac{1}{2} I (30)^2 = 450 I$. Solving for $I$ gives $I = \frac{360}{450} = 0.8 \text{ kgm}^2$.
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)
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)
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)
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 solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
(2018)
In the absence of any external torque acting on the sphere in free space, the angular momentum of the system remains conserved. Therefore, angular momentum remains constant while moment of inertia increases and angular velocity decreases.
The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} – 6\hat{k}$ at $(2, 0, -3)$, about the point $(2, -2, -2)$ is given by
(2018)
The relative position vector is $\vec{r} = (2-2)\hat{i} + (0 - (-2))\hat{j} + (-3 - (-2))\hat{k} = 2\hat{j} - \hat{k}$. Torque is $\vec{\tau} = \vec{r} \times \vec{F} = (2\hat{j} - \hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k}) = -7\hat{i} - 4\hat{j} - 8\hat{k}$.