Homework Help: Physics

Posted by Mary on Friday, October 7, 2011 at 9:37pm.

The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer) is whirled around in a circle several times and released. It then moves upward on the familiar curved path of projectile motion and eventually returns to the ground some distance away. The world record for the horizontal distance is 86.75 m, achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball was released above the ground rather than at ground level. Furthermore, assume that the ball is whirled around a circle that has a radius of 1.60 m and that its velocity at the instant of release is directed 35.9 ° above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release

Physics - drwls, Friday, October 7, 2011 at 9:54pm
The centripetal force is M*Vo^2/R

R = 1.60 m in this case. The path of the hammer is an inclined circle with that radius.

You need to know the speed at release, Vo.

You can get Vo from the distance of the world recrod throw, using
86.75 m = (Vo^2/g)*sin(2*35.9)

Vo = 29.9 m/s
Now solve for the centripetal force.

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