# Rolling Motion Question: Can someone help me out with this?

## Apr 1, 2018

1.41 m, 1,85 revolutions

#### Explanation:

Let the speed of the center of mass of the ball at the bottom of the ramp be $v$. Then its angular velocity at that point is $\omega = \frac{v}{R}$, and its kinetic energy is

$\frac{1}{2} m {v}^{2} + \frac{1}{2} I {\omega}^{2} = \frac{1}{2} \left(m + \frac{I}{R} ^ 2\right) {v}^{2}$

So, energy conservation gives

$\frac{1}{2} \left(m + \frac{I}{R} ^ 2\right) {v}^{2} = m g \left({h}_{1} - {h}_{2}\right)$

or

${v}^{2} = \frac{2 g \left({h}_{1} - {h}_{2}\right)}{1 + \frac{I}{m {R}^{2}}} = \frac{10}{7} g \left({h}_{1} - {h}_{2}\right) = \frac{10}{7} \times 10 {\text{ m"\ "s"^-2 times 0.35" m" = 5" m"^2"s}}^{2}$

(here we have used the fact that $I = \frac{2}{5} m {R}^{2}$ for a solid sphere.) So

$v = \sqrt{5} \text{ m/s}$

The time the ball takes to reach the floor from this point on is given by

t = sqrt((2h_2)/g) = sqrt((2 times 2" m")/(10" m"\ "s"^2)) = sqrt(2/5)" s"

Thus, the horizontal distance the ball travels before hitting the floor is

$\sqrt{5} \text{ m/s" times sqrt(2/5)" s" = sqrt(2)" m"~~1.41" m}$

The angle the ball revolves through in time $t$ is given by

$\theta = \omega \times t = \frac{v}{R} \times t \approx \frac{1.41}{0.12} \text{ rad}$

Thus, the number of revolutions is

$\frac{\theta}{2 \pi} \approx 1.85$