(A) Given that the wheel is initially at rest,the initial angular velocity $\omega_0 = 0$.The angular acceleration $\alpha$ produced by a constant tor

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(A) Given that the wheel is initially at rest,the initial angular velocity $\omega_0 = 0$.The angular acceleration $\alpha$ produced by a constant torque $\tau$ is given by $\alpha = \frac{\tau}{I}$.After time $t$,the angular velocity $\omega$ is given by $\omega = \omega_0 + \alpha t = 0 + \left(\frac{\tau}{I}\right)t = \frac{\tau t}{I}$.The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.Substituting the value of $\omega$,we get $K = \frac{1}{2} I \left(\frac{\tau t}{I}\right)^2 = \frac{1}{2} I \cdot \frac{\tau^2 t^2}{I^2} = \frac{\tau^2 t^2}{2 I}$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

MHT CET 2020 All MHT CET

A
$\frac{\tau^{2} t^{2}}{2 I}$
B
$\left[\frac{\tau t}{2 I}\right]$
C
$\left[\frac{\tau t}{2 I}\right]^{\frac{1}{2}}$
D
$\left[\frac{\tau t}{2 I}\right]^{2}$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

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If $I, \alpha$ and $\tau$ are the moment of inertia,angular acceleration,and torque respectively of a body rotating about an axis with angular velocity $\omega$,then:

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