jago-k1
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Homework Statement
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.
Homework Equations
mgh=(1/2)I(2h/tr)2 + (1/2)m(2h/t)2
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