Rotational Work and Energy?

Question

A tennis ball, starting from rest, rolls down the hill in the drawing. At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin-walled spherical shell, and determine the range x.

I dont have a picture of the drawing but the ball rolls down a hill, goes back up slightly then launches at a angle of 35 degrees. the point at which the ball is launches is at the same height as where the ball lands. Also the vertical height from where the ball starts rolling down the hill to the height it is launched from is 1.8m.
I know the Inertia of a hollow shell sphere is I=(2/3)mr^2

Answer 1

The first step is to determine the translational velocity when the ball becomes airborne
The way to find this most easily is conservation of energy. The KE energy of the ball will be in two components:
Translational .5*m*v^2
and rotational
.5*I*ω^2
Where I is the moment of inertia and ω is the angular velocity, which is v/r when there is no slipping.

you are correct the I=(2/3)*m*r^2
note that .5*I*ω^2=.5*(2/3)*m*r^2*v^2/r^2
which is m*v^2/3

Since the ball starts from rest, all of the energy is from the loss in PE which is m*g*h

setting up the equation of energy (note that mass divides out)
g*1.8=v^2(.5+1/3)
solve for v, g=9.81
v=21.2 m/s

Assuming the launch was at the same level as impact
the equations of motion for the flight are
x(t)=21.2*cos(35)*t
and
y(t)=21.2*sin(35)*t-.5*g*t^2
when y(t)=0, impact
t=2.48 seconds
x(2.48)=43 m, which is the range.

j

Answer 2

You mean, like the wheels of a car turning and the car moving?