Question

A solid wood door, 90.0 cm wide by 2.00 m tall has a mass of 35.0 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20.0 m/s and hits the door 60.0 cm from the hinged side. The ball rebounds with a speed of 16.0 m/s along the same line. What is the angular speed of the door just after the collision with the ball ?

Answer 1

Answer:

w = 1.14 rad/s

Explanation:

Moment of inertia ()

Moment of inertia = 35 * 0.9^2 / 3 = 9.45 kg m^2

Impulse ("An impulse is equal to the net force on the object times the time period over which this force is applied. ")

Impulse = 0.5 * [16 - (- 20)]

Impulse = 18 kg m/s

angular momentum = M of I *angular speed  w

= 9.45 w

= impulse * distance  

= 18 * 0.6

 9.45 w = 18 * 0.6

w = 1.14 rad/s