# Homework Help: Moment of Interia. Help please.

1. Nov 12, 2007

### jago-k1

1. The problem statement, all variables and given/known data
How long does it take for the hanging mass to fall the same distance 31.5 cm?
I don't know how to solve for T with everything I have. Help please.

2. Relevant equations

mgh=(1/2)I(2h/tr)2 + (1/2)m(2h/t)2

3. The attempt at a solution

Okay, previous questions are[All correct]
1.
Two masses of 150. g are suspended from a massless rod at a distance of 9.0 cm from the center. What is the moment of inertial of the two-mass system about the center of the rod?

I=m1*R^2
since they are two masses that weigh the same
I= 2(.15kg)(.09m)^2=2.43E-03 Correct

2.
If the masses rotate with an angular velocity of 2.45 rad/s, what is the rotational kinetic energy of the system?

KEr=(1/2)Iw^2 = (2.43E-03 )(2.45 rad/s)^2 = 7.29E-03 J Correct

3.
Consider the setup shown in the lab manual but with the large masses removed from the support rod. If the hanging mass is 100. g and drops a distance 31.5 cm in a time of 6.1 s, what is the moment of inertia of the support rod and shaft? The radius of the shaft is 0.50 cm.

I=mr^2(gt^2/2h - 1) = 1.446E-03 kg m^2 Correct

4.
Now two masses each of 200 g are placed on the rod at a distance of 11.0 cm from the point of rotation. What is the TOTAL moment of inertia of the masses plus rod and shaft?
Yes, Computer gets: 6.286E-03 kg m^2

I=2MR^2+Io is what you get from #3
I=2(.2kg)(.11m)^2 + Io = 6.286E-03 kg m^2 Correct

2. Nov 12, 2007

### jago-k1

BUMP if anyone can help out.