10-03-25 Torque, Angular Momentum Principle

Torque

• The torque created by a force is a measure of the twist (around a reference point) that that force exerts on the system:

• Torque gets its direction in the same way that angular momentum does: from the right-hand-rule.

Calculating Torque

• The magnitude of torque can be found by using:

• The farther from the axis the force is, the greater the torque.

• The closer $\theta$ is to $90\degree$, the greater the torque.

• Engineering interpretation:

Q10.x

A $5N$ force is applied to the end of a $0.12m$ long wrench at an angle of $30$ degrees from the vertical, as shown below.

What is the torque (around the point A) that is exerted on the wrench by this force?

$A)$ <0,0,-0.60> N \cdot m

$B)$ <0,0,-0.52> N\cdot m

$C)$ <0,0,-0.43> N\cdot m

$D)$ <0,0,-0.30> N\cdot m

$E)$ <0,0,-0.25> N\cdot m

Gather) 

$|\overrightarrow F| = 5N$

$|\overrightarrow r_A| = 0.12m$

$\theta = 30\degree$

$\overrightarrow{\tau_A}=??$

Organize)

System: Wrench

Axis: Out of the page through point A

Analyze)

$\overline {\tau_A} =$ <0,0,-1> from right hand rule

$|\overrightarrow {\tau_A}| = |\overrightarrow r_A| |\overrightarrow F| \sin(\theta_r)$

$= (0.12m)(5N)\sin(90\degree - 30\degree)$

$=0.52 N\cdot m$

The answer is B

Angular Momentum Principle

• The Angular Momentum Principle is very similar to the Momentum Principle as it can be written as (change in momentum) = (appropriate type of impulse).

• If we are only interested in the rotational angular momentum, we can relate that directly to torques relative to the center of mass of the system:

$\Delta \overrightarrow L_{rot} = \overrightarrow{\tau}_{net,CM} \Delta t$

• We will see later how changes in translational angular momentum are related to torques applied at a point other than the center of mass.

Q10.x

A brake applies a constant net torque of <12,0,-3> N\cdot m (relative to the center of mass) to a wheel that initially has a rotational angular momentum of <100, 50, 20> kg \cdot m²/s. After 3 seconds of having this torque applied, what is the rotational angular momentum of the wheel?

$A)$ <112,0,17> kg\cdot m²/s

$B)$ <136,50,11> kg\cdot m²/s

$C)$ <112,50,17> kg\cdot m²/s

$D)$ < 136,50,29> kg\cdot m²/s

$E)$ <12,0,-3> kg\cdot m²/s

Gather)

$\overrightarrow \tau_{net,CM} =$ <12,0,-3> N\cdot m

$\overrightarrow L_{rot,i} =$ <100,50,20> kg\cdot m²/s

$\Delta t = 3s$

$\overrightarrow L_{rot,f} = ??$

Organize)

System: Wheel

Axis: Through COM

Surroundings: Brake

$\Delta \overrightarrow L_{rot} = \overrightarrow \tau_{net,CM} \Delta t$

$\rightarrow \overrightarrow L_{rot,f} = \overrightarrow L_{rot,i} + \overrightarrow \tau_{net,CM} \Delta t$

$\rightarrow \overrightarrow L_{rot,f} =$ <100,50,20> kg\cdot m²/s + (<12,0,-3> N\cdot m) (3s)
$\Rightarrow \overrightarrow L_{rot,f} =$ <136,50,11> kg\cdot m²/s

The answer is B