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The instantaneous angular position of a point on a rotating wheel is given by the equation Ө (t) = 2t3 – 6t2. The torque on the wheel becomes zero at : |
t = 2 s t = 1 s t = 0.2 s t = 0.25 s |
t = 1 s |
Torque : \(\tau = I \frac{d^2\theta(t)}{dt^2} = 0 \) \(\tau = 0 \Rightarrow \frac{d^2 \theta}{dt^2} = 0\) 12 t - 12 = 0 ⇒ t = 1 s |
