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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.

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

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The relation you used applies only for moving on the top half of the hill (first 25 meters). In the second half, potential energy is only transformed into translational kinetic energy.

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  • $\begingroup$ 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. $\endgroup$
    – user40096
    Apr 11, 2016 at 13:16
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    $\begingroup$ 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). $\endgroup$
    – Gert
    Apr 11, 2016 at 14:41

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