# Ap Physics C 2002 M2?

Jun 11, 2018

a) They nicely give us the rotational inertia of a disk, so all we have to do is substitute the given values into the formula. We know that each tire has a mass of $\frac{m}{4}$ and a radius of $r$. Thus

${I}_{\text{tire}} = \frac{1}{2} \left(\frac{m}{4}\right) {r}^{2} = \frac{1}{8} {r}^{2}$

There is 4 tires on the cart, so to find the total rotational inertia we must multiply by $4$.

$4 \left(\frac{1}{8} m {r}^{2}\right) = \frac{1}{2} m {r}^{2}$

b) As soon as I see "find the height of the incline", I think of conservation of energy. Since there is rotation in the tires there will be translational AND rotational kinetic energies.

$m g h = \frac{1}{2} m {v}^{2} + \frac{1}{2} I {\omega}^{2}$

We know the value of $I$ from part $a$. The total mass of the cart is $m + 4 \left(\frac{m}{4}\right) = 2 m$

$2 m g h = \frac{1}{2} \left(2 m\right) {v}^{2} + \frac{1}{2} \left(\frac{1}{2} m {r}^{2}\right) {\omega}^{2}$

$2 m g h = m {v}^{2} + \frac{1}{4} m {r}^{2} {\omega}^{2}$

$2 g h = {v}^{2} + \frac{1}{4} {r}^{2} {w}^{2}$

We know that $\omega = \frac{v}{r}$.

$2 g h = {v}^{2} + \frac{1}{4} {r}^{2} {\left(\frac{v}{r}\right)}^{2}$

$2 g h = {v}^{2} + \frac{1}{4} {v}^{2}$

$2 g h = \frac{5}{4} {v}^{2}$

$\frac{8}{5} g h = {v}^{2}$

$v = \sqrt{\frac{8}{5} g h}$

c) Once again conservation of energy is required here. Recall the potential energy of a spring is

$\frac{1}{2} k {x}^{2}$

If we equate this to the kinetic energy of the cart, we can get the value of $x$.

$\frac{1}{2} k {x}^{2} = \frac{5}{4} m {v}^{2}$

$\frac{1}{2} k {x}^{2} = \frac{5}{4} m \left(\frac{8}{5} g h\right)$

$k {x}^{2} = 4 m g h$

$x = 2 \sqrt{\frac{m g h}{k}}$

d) An example response would be "mechanical energy was lost in the collision and as a result less was transferred to the spring, reducing the compression of the spring" .

Hopefully this helps!