Question

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill

Answer 1

Answer:

W = 0.678 rad/s  

Explanation:

Using the conservation of energy:

$E_i =E_f$

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

$\frac{1}{2}IW^2+ \frac{1}{2}mV^2 = mgh$

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I =$\frac{2}{3}mR^2$

where m is the mass and R the radius, so:

I =$\frac{2}{3}(0.426kg)(11.3m)^2$

I = 36.26 Kg*m^2

Then, replacing values on the initial equation, we get:

$\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)V^2 = (0.426kg)(9.8)(5m)$

also we know that:

V =WR

so:

$\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)W^2R^2 = (0.426kg)(9.8)(5m)$

Finally, solving for W, we get:

$W^2(\frac{1}{2}(36.26)+ \frac{1}{2}(0.426kg)(11.3m)^2) = (0.426kg)(9.8)(5m)$

W = 0.678 rad/s