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Check my work on torque and angular velocity 
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#1
Nov1905, 03:43 PM

P: 146

Hey, here's the question:
You are an athlete on the high bar in a fully extended position 180 degrees from the right horizontal. My weight: 667.08N Center of mass:80.5cm Radius of gyration: 122 cm 1)Calculate the torque at the beginning of the bar and for every 30 degrees until the rotation stops or changes direction. Assume that friction from bar produced a constant torque a 30 Nm. Diagram So from my diagram: sum(torque) at horiziontal=0 0=(F1xd) + ((30Nm)) =(667.05*0.805)30 =506.9 Nm I can tell it's angular velocity is increasing by: angular acceleration=T/I in this case a=+ve.  sum(torque) at 30degrees=0 0=(F1xd) + ((30Nm)) =(667.05)(0.805cos30)30 =435.03 Nm a=+ve, so still not slowing down or changing direction  sum(torque) at 60 degrees=0 0=(F1xd) + ((30Nm)) =(667.05)(0.805cos60)30 =238.49 Nm a=+ve still.  sum(torque) at 90 degrees=0 0=(F1xd) + ((30Nm)) =(667.05)(0)30 =30Nm Angular acc (a) a=ve, so it means a change in direction or stopping? 2)calculate the angular velocity at each position assuming that the torque from the previous position was applied for a period of 0.1s. Ok, so this is just w=@2@1/t so should I be taking the @1 as 0, or 180 degrees? if @1=0 then @2=30 degrees or 0.523rads. And I just 0.5230/0.1s = 5.23rads/s  and the next one is similar @2=60, and @1=0 so 1.04rads/0.1s or should it be 1.04/0.2s? ...etc 


#2
Nov1905, 08:22 PM

P: 146

Anyone? It would be much appreciated :p



#3
Nov2005, 12:52 PM

P: 146

Did I at least get the concept of the question right?



#4
Nov2005, 11:37 PM

P: 146

Check my work on torque and angular velocity
Just wanted to bump this up. I'll take any advice, really.



#5
Nov2105, 05:14 PM

P: 146

So...nobody knows how to do this question? or has any ideas?



#6
Nov2105, 07:49 PM

P: 146

With that many views, but no responds. Please help with this question.



#7
Nov2705, 09:22 PM

P: 146

Ok I might have figured something out,
I tried to rework it...I still get the same values for the torques. Now using cons. of mechanical energy path to find ang velocity: angular velocity for the first at 180 degrees is 0. But for the the one at 60 degrees or 30 degrees depending on which reference you use, the angular velocity is: PE=mgh (h being my center of mass) I don't understand why in your example you have 1/2 mgh?? KE=0 KERotat=1/2Iw^2, I=mk^2, I =68(1.25^2), I =106 kgm^2 ok so at that point PE=KErot mgh=1/2(106)(w^2) 68*9.81*0.805=53*w^2 537=53w^2 w=(squroot)(537/53) w=3.2 rad/sec. So...How is this value going to change overtime, because I can't see a value in the equation that will change as I change every 30 degrees. Everything stays practically the same...Except maybe h? (but that's just the value of center of mass, shouldn't change). Please do help, I really tried. 


#8
Nov2705, 11:31 PM

HW Helper
PF Gold
P: 1,123

Your exercise seems to be Torque (which you've done okay)
as a cause of angular momentum or angular velocity. (Postpone Work and PE.) Now, if the Torque you had at the beginning acted on a rigid object with your radius of gyration, what angular acceleration would there be? starting from zero angular velocity, how long would it take to travel 30 degrees (=pi/6 radians)? (recall, ½at^2) How fast (angularly!) would you be going by the time you've moved 30 degrees? Now do the same thing for the next sector (a spreadsheet might help ...) 


#9
Nov2805, 12:06 AM

P: 146

Thanks for the response: I tried it immediately:
So, to your question about angular acceleration. At that position I'd have a positive ang acceleration or one that can be calculated as: a(angular acc)=T/I, I=mk^2 (k=radius of gyration) So, 506.9Nm/(68x1.25^2)=4.77rad/s^2. Now to find the angular velocity (w) : so to move that distance or 30 degrees. I'd have to be moving a distance of 0.524 rads. Using your d=1/2at^2, or @(ang displacement)=1/2at^2 so 0.524rads*2/4.77=t^2 t=0.47s? with that time value I sub into w(ang vel)=@/t w=0.524/0.47s w=1.11 rad/s? Hope that works but I wonder where this part of the question comes in torque from the previous position was applied for a period of 0.1s.<<where do I ever use the 0.1s? 


#10
Nov2805, 12:23 AM

HW Helper
PF Gold
P: 1,123

your w = 1.11 [rad/s] is the average angular speed in that interval ...
(w_f + w_i)/2 = 1.11 [rad/s] . so , w_f = 2 w_avg  w_i = 2.22 [rad/s]. Use this for the w_i in the next interval , etc. You did better than presuming 0.1 [s] for the acceleration, you found out how long it lasted. 30 degrees is a bit coarse, but it's a sim. 


#11
Nov2805, 12:53 AM

P: 146

Ok,
hmm. I guess I could have used a(ang acc)=w(final) w(initial)/time. Ok, just let me see if I can the second one then: So with the second torque, the ang accel is 4.07rad/s^2 (which decreased, makes sense starting to slow down) For @(ang displacement) I'm now using 60degrees with respect to horizontal. I had consider using 180+60 because that is the 180 degree mark, but I found the calculation screwed up, got negatives... anyway, time I found to be 0.71s. so a(ang accel)=(w2w1)/t 4.07=(w22.2)/0.71 w2=5.09rad/s (which makes sense because ang vel still increasing as you go down.) Last question, my assumption that since torque is negative, acts in opp direction of ang velocity, this is essentially where he stops or slows down and I stop calcuations? Ok, if this is correct, I thank you. Much simpler than I had thought... 


#12
Nov2805, 01:09 AM

HW Helper
PF Gold
P: 1,123

No, if alpha = 4.07 [rad/s^2],
then .507 [rad] = 2.22 [rad/s] t + ½ 4.07 [rad/s/s] t^2 , so t is 0.194 [s] 


#13
Nov2805, 01:42 AM

P: 146

So then 4.07=(w2w1)/t (4.07*.194)+2.2=w2 w2=2.98 


#14
Nov2805, 03:15 PM

P: 146

Wait, couldn't I just use time as 0.1s for the first sec. then 0.2s for the second, then 0.3 s for the third. It'd be so much more simpler and less calcuations...
Also, one critical question. It says when it stops or changes direction, stop calculations...Doesn't that mean when the torque is smaller than the frictional force?? I can do up to 90 degrees but everything after seems a bit messy. Do I use 120 degrees for the next one or just 60 degrees again? 


#15
Nov2805, 04:08 PM

P: 146

Ok using excel, and using time values not through the quadratic method(hope that still worked) I got the folliwn values
I feel good if I'm right lol. 


#16
Nov2805, 11:43 PM

P: 146

Someone please check. I have to hand it in tomorrow morning at 9:00 or it's 10 % off my head.I really need this.



#17
Nov2905, 01:16 AM

P: 146

ah crap, half n hour before sleep...bump bump bump lol



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