Finding angular speed, angular degrees and angular acceleration?

Question

You are participating in league bowling with your friends. Time after time, you notice that your bowling ball rolls to you without slipping on the flat section of track. When the ball encounters the slope that brings it up to the ball return, it is moving at 3.91 m/s. The length of the sloped part of the track is 2.00 m. The radius of the bowling ball is 11.5 cm.

(a) What is the angular speed of the ball before it encounters the slope?

______ rad/s

(b) If the speed with which the ball emerges at the top of the incline is 0.40 m/s, what is the angle (assume constant) that the sloped section of the track makes with the horizontal?

______ °

(c) What is the magnitude of the angular acceleration of the ball while it is on the slope?

______ rad/s^2

Answer 1

In general, v=wr, with

w=angular velocity in rad/s

v=speed of the bowling ball

r=the radius of the bowling ball

Using that formula answers question a)

similarly

a=Ar

a=acceleration of the bowling ball

A=angular velocity in rad/s^2 (normally alpha is used instead of a capital A)

r= radius

Since the ball loses energy as it moves up the slope, we can calculate the height at which the ball has moved, as function of it's energy.

E=mgh=E2-E1

E2=energy when it reaches the top mv2^2/2

E1=energy when it reaches the bottom=mv1^2/2

v2=velocity at the top

v1=velocity at the bottom

m=mass of the bowling ball

g=gravitational constant 9.81 m/s^2

once you've calculated h, you can determine the angle, since we know that the sloped part is 2 meters.

sin(angle)=h/2

to calculate c), you need to calculate the component of gravity along the slope.

F=mgsin(angle).

F=ma=mAr.

so mgsin(angle)=mAr

A=gsin(angle)/r