Question
A 4.0-kg particle moves in a circle of radius 2.0 m. The angular momentum of the particle varies in time according to $l=5.0 t^{2}$. (a) What is the torque on the particle about the center of the circle at $t=3.4 \mathrm{s}$ ? $(\mathrm{b})$ What is the angular velocity of the particle at $t=3.4 \mathrm{s}$ ?
Step 1
0 t^{2}$. The rate of change of angular momentum with respect to time gives us the torque. So, we differentiate the equation with respect to time to get the torque. $\tau = \frac{dl}{dt} = \frac{d(5.0 t^{2})}{dt} = 10t$ Show more…
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