Rotational Motion Questions

The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is :

Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio :

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its midpoint and perpendicular to its length is $I_0$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is:

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta (t)=2t^3-6t^2$. The torque on the wheel becomes zero at:

A light rod of length $l$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:

A solid sphere of mass $m$ and radius $R$ is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($E_{\text{sphere}}/E_{\text{cylinder}}$) will be:

Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ ($I_B>I_A$) have equal kinetic energy of rotation. If $L_A$ and $L_B$ be their angular momenta respectively, then:

From a disc of radius $R$ and mass $M$, a circular hole of diameter $R$, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

A force $\vec{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$ 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:

Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length $L$, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point $P$ on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ${\omega }_0$ is minimum, is given by:

An automobile moves on a road with a speed of $54{\text{km h}}^{-1}$. The radius of its wheels is $0.45\text{m}$ and the moment of inertia of the wheel about its axis of rotation is $3{\text{kg m}}^2$. If the vehicle is brought to rest in $15\text{s}$, the magnitude of average torque transmitted by its brakes to the wheel is:

A small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is a:

A rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$. When the string is cut, the initial angular acceleration of the rod is:

When a mass is rotating in a plane about a fixed point, its angular momentum is directed along:

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through:

A circular platform is mounted on a frictionless vertical axle. Its radius $R=2\text{m}$ and its moment of inertia $I=200{\text{kg m}}^2$. It is initially at rest. A $50\text{kg}$ man stands on the edge and begins to walk along the edge at a speed of $1{\text{ms}}^{-1}$ relative to the ground. Time taken by the man to complete one revolution is:

A mass on a string rotates on a table. If radius is decreased by factor of $2$ by pulling the string, the kinetic energy will:

A uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A$. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $\frac{m{l}^2}{3}$, the initial angular acceleration of the rod will be:

A wheel has an angular acceleration of $3.0{\text{rad/s}}^2$ and an initial angular speed of $2.00\text{rad/s}$. In a time of $2\text{s}$, it has rotated through an angle (in radian) of:

Assertion: For a system of particles under central force field, the total angular momentum is conserved.Reason: The torque acting on such a system is zero.

A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity $v\text{m/s}$. If it is to climb the inclined surface then $v$ should be: