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Rotational kinematics and torque
Concepts tested here
- Angular acceleration
All Questions
2011 AIPMT-PRE 1 question
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The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$. The torque on the wheel becomes zero atTorque $\tau = I\alpha$, so the torque is zero when the angular acceleration $\alpha$ is zero.
$\theta = 2t^3 - 6t^2$
Angular velocity $\omega = \frac{d\theta}{dt} = 6t^2 - 12t$
Angular acceleration $\alpha = \frac{d^2\theta}{dt^2} = 12t - 12$
Setting $\alpha = 0$: $12t - 12 = 0$
t = 1 s