(C) Initial angular speed $\omega_1 = 20\ rad/s$.Final angular speed $\omega_f = 0\ rad/s$ at $t = 4\ s$.Angular retardation $\alpha = \frac{\omega_1

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(C) Initial angular speed $\omega_1 = 20\ rad/s$.Final angular speed $\omega_f = 0\ rad/s$ at $t = 4\ s$.Angular retardation $\alpha = \frac{\omega_1 - \omega_f}{t} = \frac{20 - 0}{4} = 5\ rad/s^2$.Angular speed after $t' = 2\ s$ is $\omega_2 = \omega_1 - \alpha t' = 20 - (5 \times 2) = 10\ rad/s$.Work done by the torque is equal to the change in rotational kinetic energy:$W = \Delta K = \frac{1}{2} I (\omega_1^2 - \omega_2^2)$.$W = \frac{1}{2} \times 0.20 \times (20^2 - 10^2)$.$W = 0.1 \times (400 - 100) = 0.1 \times 300 = 30\ J$.

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

Medium

$A$ 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
$10$
B
$20$
C
$30$
D
$40$

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Chapter

System of Particles and Rotational Motion

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

$A$ tangential force $F$ is applied on the rim of a ring of radius $R$. If the ring rotates by an angle $\theta$,what is the work done by the force?

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$A$ 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:

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Write the formula of work done by torque in a rotational rigid body about a fixed axis.

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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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$A$ thin rod $MN$,free to rotate in the vertical plane about the fixed end $N$,is held horizontal. When the end $M$ is released,the speed of this end,when the rod makes an angle $\alpha$ with the horizontal,will be proportional to (see figure):

DifficultJEE MAIN 2018

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$A$ 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 .......... $J$.

Similar Questions

$A$ tangential force $F$ is applied on the rim of a ring of radius $R$. If the ring rotates by an angle $\theta$,what is the work done by the force?

$A$ 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:

Write the formula of work done by torque in a rotational rigid body about a fixed axis.

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 $....... \ s$.

$A$ thin rod $MN$,free to rotate in the vertical plane about the fixed end $N$,is held horizontal. When the end $M$ is released,the speed of this end,when the rod makes an angle $\alpha$ with the horizontal,will be proportional to (see figure):

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