# Angular Velocity and Distance Problems

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A 1.62 oz golf ball with a 1.70 in. diameter is hit by a club at point P. Velocity magnitude of the ball's center of mass is 180 ft/s after the impact. What is the angular velocity of the ball? (h = 0.2 in.)

A 1 oz. bullet with a horizontal velocity of 350 ft/s strikes a circular disk. The bullet is embedded into the disk after impact. What is the angular velocity of the cylinder with the embedded bullet after impact? The disk weighs 8 lb and has a radius of 1.2 ft. Assume the bullet's mass is uniformly distributed in the disk after impact.

Please see the attachment and ignore the 3rd problem.

A 4 kg uniform rod is at rest on the frictionless horizontal x-y plane when it is struck by a 1 kg disk as shown. The coefficient of restitution of the collision is 0.6. What is the angular velocity of the rod after collision? Neglect the friction between the disk and the rod.

A bullet weighing 2 oz was fired horizontally into a 100 lb sand box suspended on a rope 3 ft long as shown. It was calculated from the observed angle that the box with the bullet embedded in it swung to a height of 0.097 ft. What was the speed of the bullet as it entered the box? Assume the box is small and acts like a particle.

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##### Solution Summary

This solution is provided in approximately 788 words. It uses linear momentum and angular momentum law to calculate angular velocity and uses energy law to calculate displacement.

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A.) Mass of the ball, m = 1.62 oz = 1.62/16 lb

radius of the ball, r = 1.7/2 in = 1.7/(2*12) ft

h = 0.2 in = 0.2/12 ft

speed of the ball after the impact, v = 180 ft/s

moment of inertia of the ball, I = (2/5)*m*r^2

Let us assume the impulse be Im, and angular velocity of the ball be, w.

Hence, by linear momentum, Im = m*v

By angular momentum

Im*h = I*w

=> w = Im/w = m*v*h/(2*m*r^2/5)

=> w = 5*v*h/(2*r^2) = 5*180*(0.2/12)/(2*(1.7/(2*12))^2)

=> w = 1494.8 rad/s ~= 1490 rad/s --Answer(b)

1.) mass of the bullet mb = 1 oz = 1/16 lb

speed of the bullet before collision vb = 350 ft/s

radius of the disk R = 1.2 ft

mass of the disk md = 8 lb

moment of inertia of the disk about O, Id = (1/2)md*R^2 + md*R^2 = (3/2)*md*R^2 = 1.5*8*1.2^2 = 1

7.28 lb.ft^2

After impact moment of inertia of the bullet about O, Ib = ...

###### Education

- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India

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