Objects with Different Moments of Inertia Rolling Down an Inclined Plane

[lc99]

Homework Statement

Homework Equations

inertia equations

The Attempt at a Solution

I think the answer for this is B) or D) but I am not sure what the sentence "a special solid cylinder in which the density is proportional to the radius" means...

The solid sphere has little inertia because mass is closer. So , solid sphere will not be last compared to the hollow sphere. The frictionless cube will not be last because most of the energy will go into KE and not RE , so the energy is not wasted. It will go down faster.

The hollow sphere will have less inertia than the hollow cylinder because the mass is closer to center of mass. Choosing between, hollow cylinder and solid cylinder... i think it might be hollow cylinder

 

[haruspex]

a special solid cylinder in which the density is propotional to the radius

It just means that the cylinder has been constructed so as to have that density distribution, with the centre less dense than the outside.

i think it might be hollow cylinder

Right... so why did you say B or D?

 

[Gene Naden]

I think you have some good insight into the problem. Not sure what your question is about the special solid cylinder.

You could use conservation of energy to figure out the time required for an object to roll down. The change in potential energy is from the vertical distance the center of mass has moved. The kinetic energy is $\frac{1}{2}I\omega^2$. The kinetic energy is equal to the change in potential energy, positive. So you get $\omega$, the angular rate of rotation, as a function of the vertical distance moved s. You have $\frac{d s}{d t} = \omega R sin(\theta)$ where R is the radius of the sphere or cylinder and $\theta$ is the angle of the incline. You can the integrate $dt = \frac{d s}{(\frac{ds}{dt})}$.

 

[Gene Naden]

The time mainly depends on $\sqrt{\frac{I}{MR^2}}$

 

[Gene Naden]

I think there is a second term in the kinetic energy, $\frac{1}{2}mV^{2} = \frac{1}{2}R^{2}\omega^2$. This adds a term to I in the final expression for the time to roll down.

 

[haruspex]

I think there is a second term in the kinetic energy, $\frac{1}{2}mV^{2} = \frac{1}{2}R^{2}\omega^2$. This adds a term to I in the final expression for the time to roll down.

It depends which I you use. If you take moment of inertia about the point of contact then you don't need to a linear term.