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Rod + 2 Particle masses = angular acceleration...

by ninjagowoowoo
Tags: acceleration, angular, masses, particle
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ninjagowoowoo
#1
Apr7-05, 09:41 PM
P: 75
PROBLEM:
A rigid rod of mass 8.70 kg and length 2.30 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m1 (mass=4.30 kg) and m2 (mass=2.60 kg) are attached at the ends of the rod. Determine the size of the angular acceleration of the system when the rod makes an angle of 50.1 with the horizontal.

I tried getting the moment of inertia using Irod = 1/12ML^2 + the Is of both the particles. Then I found the weight difference and used that to compute torque in torque = r x F. Finally I used those two answers to get angular acceleration. Of course this is wrong... can someone help me please? Or at least tell me what I did wrong... I think I may be confused about the angle between r and F. Anyway, any help would be apperciated! Thanks.
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#2
Apr7-05, 11:59 PM
P: 52
i have that problem for my physics hwk too
ninjagowoowoo
#3
Apr8-05, 11:40 AM
P: 75
haha, do you go to UCSB?

Doc Al
#4
Apr8-05, 12:14 PM
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Rod + 2 Particle masses = angular acceleration...

Quote Quote by ninjagowoowoo
I tried getting the moment of inertia using Irod = 1/12ML^2 + the Is of both the particles. Then I found the weight difference and used that to compute torque in torque = r x F. Finally I used those two answers to get angular acceleration. Of course this is wrong... can someone help me please? Or at least tell me what I did wrong... I think I may be confused about the angle between r and F.
This should work just fine. Show in detail what you did and then we can try to spot any errors.
ninjagowoowoo
#5
Apr8-05, 03:20 PM
P: 75
Irod = (1/12)ML^2 = (1/12)(8.7kg)(2.3^2)=3.83525
Iparticles = m1r1 + m2r2 = 4.3(1.15) + 2.60(1.15) = 7.935
Inet = 11.77025

Force due to gravity(net) = m1g - m2g = 42.14N - 25.48N = 16.66 N

Now this is the part where i think i may be wrong:

since the angle between r and the horizontal is 50.1 deg, then the angle between r and F is 140.1 deg.(I did try it with 50.1 deg and still no luck.)

Thus:
torque = rF(sin140.1)= 12.28953349

torque = I(alpha) = 12.28953349 = 11.77025(alpha)
so alpha = 1.04411 rad/s/s

which is wrong... is my angle wrong or is there more to it than that? Thanks for the help.
Doc Al
#6
Apr9-05, 06:29 AM
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Quote Quote by ninjagowoowoo
Irod = (1/12)ML^2 = (1/12)(8.7kg)(2.3^2)=3.83525
Iparticles = m1r1 + m2r2 = 4.3(1.15) + 2.60(1.15) = 7.935
Here's your error: the moment of inertia of a particle is [itex]mr^2[/itex], not [itex]mr[/itex].

Force due to gravity(net) = m1g - m2g = 42.14N - 25.48N = 16.66 N
Subtracting the weights of the two particles to find the net force is a shortcut that is OK here, since they happen to be equidistant from the pivot point. But each particle should be thought of as exerting its own torque, not just its own weight. You need to find the net torque, not the net force.
...
Now this is the part where i think i may be wrong:

since the angle between r and the horizontal is 50.1 deg, then the angle between r and F is 140.1 deg.
That's perfectly OK. Since you don't know whether the heavier particle is above or below the horizontal, the angle could be 140.1 deg or 39.9 deg. But it doesn't matter, the torque is the same.


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