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Rod + 2 Particle masses = angular acceleration... 
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#1
Apr705, 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. 


#2
Apr705, 11:59 PM

P: 52

i have that problem for my physics hwk too



#3
Apr805, 11:40 AM

P: 75

haha, do you go to UCSB?



#4
Apr805, 12:14 PM

Mentor
P: 41,457

Rod + 2 Particle masses = angular acceleration...



#5
Apr805, 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. 


#6
Apr905, 06:29 AM

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