Starting from a standstill (Vi = 0 m/s), a bowler bowls a bowling ball in 1.25 s

Question

Starting from a standstill (Vi = 0 m/s), a bowler bowls a bowling ball in 1.25 seconds. The bowler’s arm (shoulder to hand) is 0.75 meters in length.

a.   Just before being released, the magnitude of the hands linear velocity (Vf) is equal to 6.00 m/s. What is the magnitude of the angular velocity of the bowler’s arm at this time?

b.   What is the average angular acceleration of the bowler’s arm for the entire movement?

c.   What is the average tangential acceleration of the hand/ball for the entire movement?

Explanation / Answer

Vi= initial velocity = 0 m/s

Vf = 6m/s

t = time = 1.25 sec

length ( shoulder to arm) = radius of circular path = 0.75 m

a) w (angular velocity) = Vf / r V = w*r

w = 6 / 0.75 = 8 rad/s

b) @ ( average angular acceleration) = total change in angular velocity / total time

@ = w/ t

@ = 8 / 1.25 = 6.4 rad/s2

c) a ( average tangential acceleration) = @*r

a = 6.4*0.75= 4.8 m/s2

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