# Homework Help: Rotational kinetic energy problem

1. Jul 1, 2008

### bodensee9

Hello:

can someone help me with the following?

A rigid assembly of a thin hoop (of mass m and radius R of 0.15m) positioned on top of a thin radial rod (of mass m and length L = 2.0 R). (So imagine a circle on top of a stick, whereby if you extend the stick you will hit the center of the circle). If we give the assemblly a nudge the assembly will rotate around a horizontal axis that is perpendicular to the rod and lying on the bottom of the rod. Assuming that the energy given in the nudge is negligible, what is the assembly's angular speed about the rotation axis when it passes through the upside-down (inverted position)?

This is what I did:

I thought that since mechanical energy is conserved, so change in kinetic energy + change in potential energy = 0.

Kinetic energy = 1/2*I*v^2, where v = angular speed. I = moment of inertia, and in this case I found the inertia of the hoop = m*R^2 + m*(3R)^2 = 10*m*R^2.
The inertia of the rod = 1/3*m*L^2 = 1/3*m*(2R)^2.
Add the two up I got 34/3*m*R^2.

the potential energy I can treat as concentrated in the center of mass of the assembly. the y coordinate of the center of mass I found by: (m*R + m*3R)/2*m = 2R. Then I treat the total mass of 2m as concentrated at 2R (or -2R because it is inverted)

So my equation would give me

1/2*34/3*m*r^2*v^2 = 2R*2m*g
But I don't think this is right.

Any suggestions?

Thanks!

2. Jul 1, 2008

### Staff: Mentor

Almost right. Set the KE equal to the (negative of the) change in PE. (What's the change in PE?)

3. Jul 1, 2008

### bodensee9

Oh okay, I think the displacement would be 4R in that case? Thanks!

4. Jul 1, 2008

Right!

5. Mar 7, 2011

### Nash77

I understand the general equations in this problem, but why is the hoop being multiplied by (3R) [I = mR^2 + m(3R)^2] and the rod being multiplied by (1/3) [I = 1/3mL^2

6. Mar 7, 2011

### Nash77

Kinetic energy = 1/2*I*v^2, where v = angular speed. I = moment of inertia, and in this case I found the inertia of the hoop = m*R^2 + m*(3R)^2 = 10*m*R^2.
The inertia of the rod = 1/3*m*L^2 = 1/3*m*(2R)^2.

^^ I understand the moment of inertia equations, I = mR^2, but why are we multiplying by 3 for the hoop and by 1/3 for the rod?

7. Mar 8, 2011

### Staff: Mentor

For the hoop: The rotational inertia of a hoop about its center of mass is I = mR^2; here we have to use the parallel axis theorem to find the rotational inertia about the axis of rotation, which is at the end of the rod. That's where the additional term of m(3R)^2 comes from.

Note that all points of the hoop are equidistant from the center, which is what allows us to use I = mR^2. That's not the case for a rod. The formula for the rotational inertia of a rod about one end is 1/3mL^2, where L is the length of the rod.

See: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi"

Last edited by a moderator: Apr 25, 2017