# What is the translational velocity of a sphere after rolling down a hill?

A solid, uniform, spherical boulder starts from rest and rolls down a 50.0-m-high hill. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill?

The answer is 29.0 m/s.

This is what I did:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ Substituting $I$ for a solid sphere and using $v = \omega r$... $$mg(50m) = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$$ $$50g = \frac{7}{10}v^2$$ $$v = 26.5 m/s$$

However, this is not the answer, and I'm not sure where I went wrong.

• You should add the radius of the sphere to the height, because the potential energy is measured for the center of the mass. Apr 11, 2016 at 12:37
• The radius of the sphere appears to not be given though.
– user40096
Apr 11, 2016 at 13:13

• I see. So $mg(25m) = \frac{7}{10}mv^2$, so $v=18.7 m/s$ then the rest of the hill is $mg(25m) = \frac{1}{2}mv^2$, so $v = 22.13 m/s$, thus final velocity is the sum $18.7 m/s + 22.13 m/s=40.8 m/s$? This does not seem correct.
• You can't just add up the velocities. At the end of the first 25 m the bolder has a velocity of 18.7 m/s. In the second 25 m its rotational energy doesn't change any more but its translational energy does. $mgh=\Delta K$., So $mgh=\frac12 m(v_2^2-v_1^2)$, where $h=25$ and $v_1=18.7$. Then calculate $v_2$ as the final velocity (28.98 m/s).