Question

A ball in the shape of a uniform spherical shell (like a soccer ball) of mass 1.5 kg and radius 15 cm rolls down a 35° incline that is 6.0 m high, measured vertically. The ball starts from rest, and there is enough friction on the incline to prevent slipping of the ball.

(a) How fast is the ball moving forward when it reaches the bottom of the incline, and what is its angular speed at that instant?

(b) If there were no friction on the incline, how fast would the ball be moving forward and what would be its angular speed at the bottom?

Answer 1

a) We apply the conservation of energy:

$E_o =E_f$

$mgh = \frac{1}{2}mv^2 +\frac{1}{2}I\omega^2$

$mgh = \frac{1}{2}mv^2 +\frac{1}{2}\left ( \frac{2}{3}mr^2 \right )\left ( \frac{v}{r} \right )^2$

solving for v we get:

$v= \sqrt{\frac{6}{5}gh} = \sqrt{\frac{6}{5}(9.8)(6)} = 8.4 m/s$

then the angular speed is:

$\omega = v/r = 8.4/0.15 = 56 rad/s$

b.When there is no friction, the ball is slipping so the angular speed is zero.

$E_o = E_F \rightarrow mgh = mv^2/2$7

$v = \sqrt{2gh} = \sqrt{2*9.8*0.15}= 1.715 m/s$

$\omega = 0$