Two Problems on Rotation of a Rigid Body

[mst3kjunkie]

the first

Homework Statement

Calculate the moment of inertia about a center axis of a long rod with an LxL square cross section

Homework Equations

moment of inertia (I) = the integral of the radius^2 dm

The Attempt at a Solution

I'm not really sure where to start with this one, but the I of a thin rod about its center is equal to 1/12 ML^2. This however, has a cross-section. How do I take this into account and solve the problem?

and the second:

Homework Statement

A Hollow sphere is rolling along a horizontal floor at 5.0 m/s when it comes to a 30 degree incline. How far up the incline does it roll before coming back down?

omega (angular velocity) before incline=5.0 m/s
omega at top point on incline before rolling back=0m/s

Homework Equations

Moment of Inertia of a hollow sphere= 2/3MR^2

The Attempt at a Solution

once again, I'm not sure where to begin or how to set this one up, but I think if I could figure this out I could solve the problem.

 

[radou]

5.0 m/s cannot be angular velocity, look at the units. It is the translatoral velocity of the sphere's center of mass, I assume.

 

[mst3kjunkie]

that was a mistake on my part. I meant to put velocity, not angular velocity.

 

[radou]

that was a mistake on my part. I meant to put velocity, not angular velocity.

Ok, now all you have to do is use energy conservation. Note only that the sphere has translatoral and rotational kinetic energy.

 

[mst3kjunkie]

I'm a bit unsure as to how to set up the formula, still.

 

[radou]

I'm a bit unsure as to how to set up the formula, still.

Do you know how to use conservation of energy? The kinetic energy (translatoral + rotational) of the sphere must equal the potential energy of the sphere at the highest point on the incline (since the sphere's kinetic energy equals zero at that point).

 

[mst3kjunkie]

I've used Conservation of Energy before, but I haven't encountered the translatoral (at least the term) or rotational energy formulas yet.

 

[radou]

I've used Conservation of Energy before, but I haven't encountered the translatoral (at least the term) or rotational energy formulas yet.

$$E_{K,T}=\frac{1}{2}mv^2$$, and
$$E_{K,R}=\frac{1}{2}I \omega^2$$.

 

[mst3kjunkie]

okay, now I"ve gotten it to

(25/2)m+(1/3)mr^2(omega)^2 = 9.8mh

am I on the right track?

 

[radou]

Yes, but note only that the moment of inertia of the sphere must be calculated with respect to the bottom point, since the translational velocity at that point equals zero, so it is the center of rotation. This link may be a useufl reference: http://www.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html" . I hope you'll get everything right now.