Question

The hammer throw is an Olympic event where a 7.5 kg metal ball is attached to a 1.2 m long wire, as shown in the diagram. One such competitor was observed to swing the hammer around in a horizontal circle with an average angular acceleration a = 5.6 rad/s?. After 3 seconds the hammer was released. (a) Calculate the angular velocity and tangential speed of the metal ball just before it is released. (b) Calculate the number of revolutions the metal ball undertook before being released. (c) Calculate the centripetal force applied to the ball and hence the tension in the wire just before the hammer is released.

Answer 1

Solution

a)

We know,

10=

w=5.6 * 3 rad) s

w = 16.8 rad/s

tangential speed

v = rw = 1.2 * 16.8 m/s = 20.16 m/s

b)

We know,

W2 = 200

5 Ꮎ =

16.82 2 *5.6 rad

A= 25.2 rad

one rev =$2\pi$ rad

Therefore 25.2 rad = 4 revolutions approx

c)

Centripetal force = mv²/r = 7.5 * 20.16² / 1.2 = 2540.16 N

Since the hammer was in horizontal motion tension = centripetal force = 2540.16 N

If any doubt feel free to comment