(C) The angular displacement is given by $\theta(t) = 2t^3 + 0.5$.Angular velocity $\omega$ is defined as the rate of change of angular displacement w

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(C) The angular displacement is given by $\theta(t) = 2t^3 + 0.5$.Angular velocity $\omega$ is defined as the rate of change of angular displacement with respect to time: $\omega = \frac{d\theta}{dt}$.Differentiating $\theta$ with respect to $t$: $\omega = \frac{d}{dt}(2t^3 + 0.5) = 6t^2$.To find the angular velocity at $t = 2 \ s$,substitute $t = 2$ into the expression for $\omega$:$\omega = 6(2)^2 = 6 \times 4 = 24 \ rad/s$.

Class 11 Physics 2 Motion in a Plane Kinematics Circular Motion (Uniform Angular Accelaration)

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If the equation for the angular displacement of a particle moving on a circular path is given by $\theta = 2t^3 + 0.5$,where $\theta$ is in radians and $t$ is in seconds,then the angular velocity of the particle after $2 \ s$ from its start is ......... $rad/s$.

AIIMS 1998 All AIIMS

A
$8$
B
$12$
C
$24$
D
$36$

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If the equation for the angular displacement of a particle moving on a circular path is given by $\theta = 2t^3 + 0.5$,where $\theta$ is in radians and $t$ is in seconds,then the angular velocity of the particle after $2 \ s$ from its start is ......... $rad/s$.

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