# Torque Self-Test: Definition of Torque

Several children are playing in the park. One child pushes the merry-go-round with a force of 50 N. The diameter of the merry-go-round is 3.0 m. What torque does the child apply? Recall \(\tau = r \times F.\).

The child applied a torque of:

- A. 70 N m into the screen
- B. 70 N m out of the screen
- C. 141 N m into the screen
- D. 141 N m out of the screen

**A. Correct**- B. No. Remember that it is F rotated into r
- C. No. Remember that it is F rotated into r
- D. No. Remember that it is F rotated into r

Recall that torque, \(\tau\) , is the cross product of the moment arm, r, and the force applied, F. In other words \(\tau = rF \sin \theta\)

Since the pivot point (the point about which the object rotates) is at the centre of the merry-go-round, then the moment arm vector, r, points from the centre to the rim of the merry-go-round. Using the Right Hand Rule, we find that the torque points into the screen.

Be careful with the moment arm vector. It is not the diameter of the merry-go-round, which was the given quantity, but the radius. So

\(r = d/2 = 1.5 m\)

Also, note that the angle between the two vectors, placed tail-to-tail, would still be \(\theta = 110 ^\circ\). Substituting for r, F, and \(\theta\), we get

\(\tau = (1.5) (50) \sin(110)\)

\(\tau = 70 N \) (into the screen)