8. You roll a 16 pound bowling ball down the lane at 8 meters per second. Assume no slippage and ignore friction and air resistance. The radius of the bowling ball is 4.25 inches. a. What is Ω (the rotation rate in radians per second)? b. What is the angular momentum? c. What is the rotational energy? d. As you look down the lane, in which direction does the angular momentum vector point?

A bowling ball weighing 16 pound rolls down the lane at 8 m/s. The ball does not slip, and friction, as well as air resistance, can be ignored.
The mass of the ball is 16 pound which is equivalent to 7.25748 kg. The linear velocity of the ball is 8 m/s. The radius of the ball is 4.25 inches or 0.10795 m.
Linear velocity v and rotational velocity w are related by w = v/r.
Substituting the values given,
w = 8/0.10795 = 74.1 rad/s
The angular momentum of a body is given by L = I*w, where I is the moment of inertia and w is the angular speed. For a solid spherical body of mass m and radius r, moment of inertia I = (2/5)*m*r^2
The moment of inertia of the bowling ball is I = (2/5)*7.25748*(0.10795)^2 = 0.03383 kg*m^2
The angular momentum is equal to 0.03383*74.1 = 2.507 kg m^2/s
The rotational energy of the ball is (1/2)*I*w^2 = (1/2)*.03383*74.1^2 = 92.877 J
The right hand rule gives us the direction of the angular momentum vector. It is pointing to the left.