Question

Question 1-4: A golf club 1.01 m long completes a downswing in 0.25 s through a range of 180 degrees with a uniform angular velocity. 1. What is the angular velocity of the club? A. 12.57 rad/s B. 550 °/s C. 11.3 rad/s D. 659 °/s 2. What is the linear distance moved by the end of the club? A. 181.8 m B. 172.6 m C. 2.45 m D. 3.17 m 3. What is the tangential velocity of the end of the club? A. 12.7 m/s B. 13.6 m/s C. 11.6 m/s D. 3.2 m/s 4. What is the tangential acceleration of the end of the club? A. 101.5 m/s2 B. 100.5 m/s2 C. 50.8 m/s2 D. 0 m/s2

Answer 1

1)t = 0.25 s

w = pi/4 = 3.14/0.25 = 12.56 rad/s

Hence, w = 12.56 rad/s

Linear distance will be equal to

D = pi L

D = 3.14 x 1.01 = 3.17 m

Hence, D = 3.17 m

the velocity will be:

v = r w = L w

v = 1.01 x 12.56 = 12.7 m/s

Hence, v = 12.7 m/s

a = 0 m/s^2

since it moves in a circle and has its origin fixed.