Question

An object's angular momentum changes by 10 kg-m2/s in 2.0 s. What magnitude average torque acted on this object? *

Answer 1

Answer:

$\tau=5N\cdot m$

Explanation:

The first derivative of the angular momentum with respect to time is defined as the magnitude of the average torque acting on the particle:

$\frac{dL}{dt}=\tau$

So, we calculate the magnitude of the average torque, dividing the change of the angular momemtum into the time:

$\tau=\frac{\Delta L}{\Delta t}\\\tau=\frac{10\frac{kg\cdot m^2}{s}}{2s}\\\tau=5N\cdot m$