(D) When a body rolls without slipping,it possesses both translational kinetic energy and rotational kinetic energy.Translational kinetic energy is gi

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$ \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} $

(D) When a body rolls without slipping,it possesses both translational kinetic energy and rotational kinetic energy.Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the center of mass.Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia about the center of mass and $\omega$ is the angular velocity.The total kinetic energy $K$ is the sum of these two energies:$K = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Therefore,the correct option is $D$.

Class 11 Physics System of Particles and Rotational Motion Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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$A$ sphere of mass $m$ and moment of inertia $I$ about its center of mass is rolling without slipping down an inclined plane from rest. Which of the following represents its total kinetic energy?

A
$ \frac{1}{2}I{\omega ^2} $
B
$ \frac{1}{2}m{v^2} $
C
$ I\omega + mv $
D
$ \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} $

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System of Particles and Rotational Motion

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Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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The $M.I.$ of a body about a given axis is $1.2 \ kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied about that axis for a time duration of $....... \ s$.

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$A$ sphere of mass $m$ and moment of inertia $I$ about its center of mass is rolling without slipping down an inclined plane from rest. Which of the following represents its total kinetic energy?

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The $M.I.$ of a body about the given axis is $1.2 \, kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \, J$,an angular acceleration of $25 \, rad/s^2$ must be applied about that axis for a duration of ........ $s$.

$A$ thin rod of length $l$ and mass $m$ oscillates in a vertical plane about a horizontal axis passing through one of its ends. If the maximum angular velocity of the rod is $\omega$,what is the maximum height reached by its center of mass?

$A$ wheel is rotating with an angular speed of $20\ rad/s$. It is stopped to rest by applying a constant torque in $4\ s$. If the moment of inertia of the wheel about its axis is $0.20\ kg\cdot m^2$,then the work done by the torque in two seconds will be .......... $J$.

$A$ thin rod of mass $m$ and length $l$ is oscillating about a horizontal axis through its one end. Its maximum angular speed is $\omega$. Its centre of mass will rise up to a maximum height of:

The $M.I.$ of a body about a given axis is $1.2 \ kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied about that axis for a time duration of $....... \ s$.

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