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:

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

Answer 2

The magnitude of average torque acting on the object is 5.0 Newton-meter.

Given the data in the question;

• Angular momentum; $L = 10kgm^2/s$

• Time elapsed; $t = 2.0s$

• Magnitude average torque; $T = \ ?$

Torque

Torque is simply the time rate of change of angular momentum.

This is expressed as;

$T = \frac{\delta l}{\delta t}$

Where $\delta l$ is change in angular momentum and $\delta t$ is change in time.

To determine the magnitude of average torque acting the object, we substitute our given into the expression above.

$T = \frac{\delta l}{ \delta t} \\\\T = \frac{10kgm^2/s}{2.0s}\\ \\T = 5.0kgm^2/s^2\\\\T = 5.0kgm.m/s^2\\\\T = 5.0N.m$

Therefore, the magnitude of average torque acting on the object is 5.0 Newton-meter.

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