# 0.32 Phy1320: angular momentum -- the mathematics of torque  (Page 12/13)

 Page 12 / 13

The torque

Referring back to Figure 6 , we find that the torque produced by this force is equal to the product of the distance from the center to P and the tangential orperpendicular component of the force vector.

Therefore,

T = Ft*r, or

T = F*sin(A)*r

where

• T represents torque
• Ft represents the tangential component of the force vector
• r represents the distance from the puck to the center of the circle
• F represents the force vector
• A represents the angle between the force vector and the line from the puck to the center of the circle

## Example scenarios

This section contains several example scenarios involving torque.

## Net torque on the Lazy Susan turntable

The turntable discussed earlier, which has a radius of 24 cm, is spinning clockwise. You press your fingers on the east and west sides of the turntable with equalforces of 6.67 N. The coefficient of friction between the turntable and your fingers is 0.75. What is the net torque on the wheel?

Solution:

The 13.3 N force on each side of the table creates a tangential kinetic friction force on each sideof the table equal to

Ft = 6.67 N * 0.75

Each force is in the opposite direction of the direction of rotation.

The net torque is equal to the sum of the torques.

Each torque is equal to the product of the force and the distance from the center of the turntable to the point at which the force is applied.

T = 2 * Ft * r, or

T = 2 * 6.67 newtons * 0.75 * 24 cm

Entering this expression into the Google calculator gives us

T = 2.4 joules

However, this is one case where the Google calculator gives us a misleading answer. We know that torque is not measured in joules. Instead, torque ismeasured in N*m. Therefore, the net torque on the turntable is

T = 2.4 N*m

## A door-closing scenario

When viewed from above, the scenario is a door that is open. From above, the wall to which the door is attached can be represented by a horizontal line that runs from west to east. The door can be represented by a line segment at an angle of about 45 degreessouth of east. The line segment (door) is attached to the wall at the upper-left end of the line segment. That is the point where the door is hinged, and that point is the axis ofrotation for the door.

Assume that the axis of rotation extends out of the page towards you.

The door will need to rotate about 45 degrees counter clockwise to become flush with the wall and be closed.

A person is standing on the north side of the wall pulling on a rope that is attached to the door. The rope is attached 11.5 cm from the hinge and makes a 45degree angle with the surface of the door. That person pulls on the rope with a force of 51 N.

Using the door hinges as the axis of rotation, find the magnitude of the torque that is exerted on the door. What is the sign of the torque.

Solution:

This solution is based on the cross product from Figure 4 .

The magnitude of the torque is given by

T = r * F * sin(angle), or

T = 11.5 cm * 51 newtons * sin(45 degrees)

Entering this expression into the Google calculator gives us

T = 4.15 N*m

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