
#1
Nov2309, 09:56 PM

P: 21

1. The problem statement, all variables and given/known data
A student stands on a platform that is free to rotate and holds two dumbbells, each at a distance of 65 cm from his central axis. Another student gives him a push and starts the system of student, dumbbells, and platform rotating at 0.50 rev/s. The student on the platform then pulls the dumbbells in close to his chest so that they are each 22 cm from his central axis. Each dumbbell has a mass of 1.00 kg and the rotational inertia of the student, platform, and dumbbells is initially 2.40 kg•m2. Model each arm as a uniform rod of mass 3.00 kg with one end at the central axis; the length of the arm is initially 65 cm and then is reduced to 22 cm. What is his new rate of rotation? 2. Relevant equations The initial angular momentum: L=Iw The moment of inertia of a rod about one end: =1/3 m L² The moment of inertia of the weight: =m L² 3. The attempt at a solution To solve this problem, use conservation of angular momentum. The initial angular momentum is L = I ω I is the initial moment of inertia=2.40 kgm² ω is the initial angular velocity 1.15 rev / sec converted to into radians per second= 3.14 rad/s. ****Q: Should I solve for L now? Now we need to know the change in moment of inertia. The moment of inertia of a rod about one end is 1/3 m L² 1/3(3kg)(.65m)^2=.4225kg*m multiplied by 2,for two arms =.845kg*m ****Q:Is the mass 3 kg? In this equation does "L" mean length or Initial angular momentum? The moment of inertia of the weight is simply mL² =(3kg)(.22m)^2=.1452kg*m The difference is the change in the system moment of inertia. >845kg*m.1452kg*m=.6998 kg*m I used this to compute the new moment of inertia, then use conservation of angular momentum to find the new angular velocity. L=Iw *********I'm stuck here because I am not getting the right answer when I solve for w or when I just solve for L. Not sure why I am not getting 1.3 rev/s,just wondering is there a calculation error or equation error? The answer was 1.3rev/s Thank you 



#2
Nov2309, 10:28 PM

HW Helper
P: 2,156

Also, I think you've done the conversion from revolutions per second to radians per second incorrectly. [tex]I = \frac{1}{3}ml^2[/tex] And the mass is 3kg... why would you think it would be otherwise? Note that you've computed the moment of inertia incorrectly as well  the units are wrong. 



#3
Nov2309, 11:12 PM

P: 21

"In I = 1/3mL^2, the L is length. I would write it as lowercase l to avoid confusion:
LaTeX Code: I = \\frac{1}{3}ml^2 And the mass is 3kg... why would you think it would be otherwise? Note that you've computed the moment of inertia incorrectly as well  the units are wrong." Thanks, so I=1/3(3kg)(.65^2m)=.4225 kg*m^2 Q: For one arm,now do I multiply.4225 kg*m^2 by 2 in order to take in account the second arm?If not what do I plug into the second equation for the I=1/3ml^2? Then for The moment of inertia of the weight is simply mL² =(3kg)(.22m)^2=.1452kg*m^2 same here do I multiply.1452kg*m^2 by 2? ****I would like to use the option of computing the change in moment of inertia of the weights and the change in moment of inertia of the arms, then add the two changes together. But once I manage to find the total moment of Inertia then do I use the L=Iw? Do I solved for L or w? I appreciate it the help.thanks alot 



#4
Nov2409, 03:26 PM

P: 21

Question regarding angular momentum and velocity.As it relates to a new rate rotation
Can someone give me an idea?




#5
Nov2409, 04:30 PM

HW Helper
P: 2,156

[tex]L_i = I_i\omega_i[/tex] and the final angular momentum, [tex]L_f = I_f\omega_f[/tex] are equal, [tex]L_i = L_f[/tex] and combine that with the fact that [tex]I_f  I_i = \text{change in }I[/tex] (where "change in I" is what you figured out before) Remember that you know [itex]\omega_i[/itex] and you're trying to find [itex]\omega_f[/itex]. 



#6
Nov2409, 07:35 PM

P: 21

Ok,
Still not getting th right answer: I did Inertia Arms I=1/3(3kg)(.65^2m)=.4225 kg*m^ X 2= .845 kg*m^2 Inertia Weight I=(3kg)(.22m)^2=.1452kg*m^2 X 2= .2904 kg*m^2 .845 kg*m^2+.2904 kg*m^2=1.13 kg*m^2 Once you have computed the change in moment of inertia, use the fact that angular momentum is conserved. Specifically, the initial angular momentum, LaTeX Code: L_i = I_i\\omega_i and the final angular momentum, LaTeX Code: L_f = I_f\\omega_f are equal, LaTeX Code: L_i = L_f and combine that with the fact that LaTeX Code: I_f  I_i = \\text{change in }I (where "change in I" is what you figured out before) Remember that you know LaTeX Code: \\omega_i and you're trying to find LaTeX Code: \\omega_f . Ok So I have tried to incorporate the 1.13 kg*m^2 by subtracting it from 2.40 kgm^2. But no hope. Its hard to try and figure out when a simple miscalculation may keep giving me the wrong answer. Can you show me the steps I have gotten frustrated in just trying to understand it and figure it out? When you work it out are you getting 1.3 rev/s? Thanks you have been alot of help 


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