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

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

Class 11 Physics 3-2.Motion in Plane Kinematics Circular Motion (Uniform Angular Accelaration)

Medium

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 at $t = 2 \, s$ is:

A
$8 \, rad/s$
B
$12 \, rad/s$
C
$24 \, rad/s$
D
$36 \, rad/s$

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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 at $t = 2 \, s$ is:

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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 at $t = 2 \, s$ is: