Question

I have no idea as to how to complete this problem. Can someone please draw what the problem is talking about and then help me?  

A person is in the middle of bowling. Assume the bowling motion only involves rotation at the shoulder joint (the elbow and wrist are fixed), and the trunk remains upright (vertical) the entire time. The motion begins with the shoulder in 45° of extension, and the ball is released at 45° of shoulder flexion. Solve the following problems:

a. What is the angular displacement at the shoulder joint?

b. If the distance from the shoulder joint to the center of the ball is 65 cm, what is the distance that the ball travels from start to finish of this motion?

c. If the entire motion takes 0.3 s, what is the angular velocity of the shoulder?

d. If the answer to (c) is the angular velocity of the shoulder as the ball is being released, and the shoulder is not moving at the start of the motion, what is the angular acceleration at the shoulder during this movement?

e. What is the linear velocity of the ball at the time of release?

f. What is the linear (resultant) acceleration of the ball at the time of release?

Answer 1

a) angular displacement, theta = 45 + 45

= 90 degrees

= 90*2*pi/360

= 1.57 radians

b) distance travelled by the ball, s = r*theta

= 0.65*1.57

= 1.02 m

c) w = theta/t

= 1.57/0.3

= 5.23 rad/s

d) angular acceleration, alfa = (wf - wi)/t

= (5.23 - 0)/0.3

= 17.4 rad/s^2

e) linear velocity of the ball at the time of the release, v = r*w

= 0.65*5.23

= 3.40 m/s

f) tangential acceleration, a_tan = 17.4 rad/s

radial acceleration, a_rad = r*w^2

= 0.65*5.23^2

= 17.8 m/s^2

a_total = sqrt(a_tan^2 + a_rad^2)

= sqrt(17.4^2 + 17.8^2)

= 24.9 m/s^2