# Kinetic Energy And Rotational Motion

The problem is, "A metal can containing condensed mushroom soup has mass 220 g, height 11.0 cm and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at 30.0° to the horizontal and is then released to roll straight down. It reaches the bottom of the incline after 1.50 s.

(a) Assuming mechanical energy conservation, calculate the moment of inertia of the can.

(b) Which pieces of data, if any, are unnecessary for calculating the solution?

My attempt at solving (a):

I figured that I could use the equation $\Sigma W=\Delta K=1/2I\omega_f^2-1/2I\omega_i^2$ Since the force of gravity that is acting along the incline is applied constantly over a distance, $W_g=mg\cos(60^{\circ})(3.00~m)$; and since the can rolls down the incline in 1.50 s, $v=3.00/1.5 \implies 2~m/s$, which means that $\omega_f=2/0.0319 \implies 62.695925~rad/s$ With this, and knowing that $\omega_i=0$, $mg\cos(60^{\circ})(3.00)=1/2I(62.695925)^2 \implies I=\frac{(6.00)mg\cos(60^{\circ})}{(62.695925)^2}$ When I calculated this, I got $I=0.00165~kg\cdot m^2$; however, the true answer is $I=0.000187~kg\cdot m^2$ I've re-worked my solution several times, what am I doing incorrectly?

As for (b), the answer is that the height of the can is an irrelevant piece of information, why is that?

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1) The first thing I notice is that you have stated that the velocity at the end of the ramp is $2\textrm{ m/s}$. Remember that the can is accelerating as it rolls down the ramp, so the equation $v=\textrm{d}s/\textrm{d}t$ is not applicable here for finding the instantaneous velocity at the bottom. The can does indeed average $2\textrm{ m/s}$ during its trip, but this is not the final velocity of the can. Use this new corrected value to calculate angular frequency.

2) I find this problem simpler to solve using energy analysis. Take the can's initial potential energy:

$$E_\textrm{pot} = m g h = 3.234\textrm{ J}\quad.$$

We also know that the final kinetic energy of the can must equal this due to the conservation of energy, but the final energy of the can must be broken into translational kinetic energy (due to the can's movement) and rotational kinetic energy (due to the rotation). (This is why your solution above was giving incorrect answers as it didn't take translational kinetic energy into account.) Thus, we also know that:

$$E_\textrm{pot}(t=0) = E_\textrm{kin,trans}(t=1.5\textrm{ s}) + E_\textrm{kin,rot}(t=1.5\textrm{ s})\quad,$$

which, for our case, is

$$3.234\textrm{ J} = \frac{1}{2} mv^2 + \frac{1}{2}Iω^2\quad.$$

Plugging in known values to this equation with the correct value for angular velocity $\vec \omega = \vec r \times \vec v$ gives the accepted answer:

$$I = 0.000187\textrm{ kgm}^2\quad.$$

As for part B, the height of the can is irrelevant because as long as we know the mass and radius of the can, we can solve the problem. The ‘extra mass’ resulting from lengthening the can would be centered about the can's original center of mass, and as such the moment of inertia would not be affected for this problem.

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So, the only problem I had with my approach is the miscalculation of the translational velocity? Would I still have to use a force approach to find the acceleration of the can? –  Mack Dec 3 '12 at 13:35
You shouldn't have to use a force approach to calculate acceleration, no. You can use the fact that the average velocity of the can was 2 m/s and that it accelerated at a constant rate over 1.5 seconds. Thus, because the can started from 0 m/s, its final velocity would be 4 m/s such that it averages 2 m/s for the trip. –  Mik Cox Dec 3 '12 at 23:27