Question

Q4. Tiger Woods is known for having one of the fastest driving swings in Golf (among other things). In fact, his Tee-off ball speed was recorded to be 122 MPH.

1. Determine what Tiger Wood’s angular velocity and acceleration are for the club. He is 1.85 meters tall. (Refer to the Anthropometry slides to estimate his shoulder height)

Shoulder height = 1.5133

122 MPH = 54.54 m/s

Answer 1

a)
Linear velocity, v = 54.54 m/s
Shoulder height, h = 1.5133 m
Angular velocity, = v/h
= 54.54 / 1.5133
= 36.04 rad/s

b)
Consider that he made a half revoulution during the swing.
Angular displacement, = rad
Initial angular velocity, i = 0
Final angular velocity, f = 36.04 rad/s
Consider the angular acceleration as

Using the formula f2 - i2 = 2
Substituting values,
2 * * = (36.04)2 - 0
= 206.7 rad/s2