A football punter hits a 0.20-kg ball at an angle of 45 above the horizontal. Th

Question

A football punter hits a 0.20-kg ball at an angle of 45 above the horizontal. The ball rises to a maximum height of 13 m from the starting point. Initially the ball is at rest. The length of the punter's leg is 0.95 m and the rotational inertia of the leg is 4.0 kgm2. Assume that the collision between the ball and the leg is elastic, such that the kinetic energy and rotational momentum of the leg-ball system are constant during the collision. Suppose that the leg accelerates from rest through a 90angle with a constant the rotational acceleration. Ignore air resistance.

Determine the rotational acceleration of the leg of the punter.

Determine the tangential acceleration of the foot of the punter

Explanation / Answer

Let angular speed of leg, just before and after strike be w and w'.
Velocity of ball in vertical direction = root(2gh)
Speed of ball = Vy/sin45 = 2 root(gh) = 22.6 m/s
Initial angular momentum of leg = Iw = 4w
final angular momentum of leg = 4w'
Anular momentum of ball = MVL = 4.3 kgm2/sec
By conservation of angular momentum
4w = 4 w' + 4.3 ....1
As collision is elastic, relative velocity between foot and ball, does not change.
Hence wL = V - w'L
0.95 w = 22.6 - 0.95w'   ....2
sloving 1 and 2 we get
w = 12.4 rad/sec
w2 = 2 alpha theta = 2 alpha (pi/2) = pi alpha
alpha = w2/pi = 48.9 rad/sec2
tangential acceleration of foot = alpha L = 46.5 m/sec2

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