
#1
Nov2311, 11:11 AM

P: 5

1. The problem statement, all variables and given/known data
Here's the question: Imagine that you are a gymnast on a high bar, 180 degrees from the right horizontal in a fully extended position (NO angular velocity at this point). Assuming that the center of mass of your body is 1.288m from your extended hands, and that your radius of gyration is 1.4m from your extended hands.. Calculate the torque created about the bar beginning at the start and for every 30 degrees until the rotation stops or changes direction. Assume that the friction from the bar in your hands produces a constant torque value of 30Nm. My mass = 91.6kg or 898.6N 3. The attempt at a solution Okay, so.. I know that T = Fxd, but I don't know which force or distance to use. Do I use F=898.6N, which is my weight? And my center of mass which is 1.288m? Also, I'm not really sure how to solve for the different angles that I (the gymnast) am rotating around the bar at. I found a similar question to mine on a different forum, and here's how this person solved it: (I plugged in my values instead of his) sum of torque at the start (0 degrees)= 0 0=(898.6*1.288)+(30Nm) =1127.392Nm  sum of torque at 30 degrees from horizontal = 0 0=(898.6*(1.288cos30))+(30Nm) =972.33Nm  sum of torque at 60 degrees from horizontal = 0 0=(898.6*(1.288cos60))+(30Nm) =548.7Nm  sum of torque at 90 degrees from horizontal = 0 0=(898.6*(1.288cos90))+(30Nm) =30Nm ..so this is the point where the gymnast changes direction and starts ascending back up the other way. I just don't understand why this person used cos in his equations and how I am supposed to break up the different components of this problem to then create a triangle that I can solve from.. so I have all of my answers, thanks to the post from this other forum, but I have no idea how he got to this point. Any help at all would be appreciated! 



#2
Nov2311, 12:11 PM

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#3
Nov2311, 08:37 PM

P: 5

okay, I completely agree with you that my notation was wrong..
so I drew out a diagram and I now understand where the use of cosine came from and what values I was looking for, but there is a part 2 to this question..  Calculate the angular velocity at each position assuming that the torque from the previous position was applied for a period of 0.1 seconds. so my torque values are as follows: at 0 degrees from horizontal, T = 1127.392Nm at 30 degrees from horizontal, T= 972.33Nm at 60 degrees from horizontal, T = 548.696Nm and at 90 degrees from horizontal, T= 30Nm so to find w (angular velocity), I attempted to first find a (angular acceleration), since a=Δw/Δt I found a from T(torque)/I(moment of inertia) My moment of inertia (I) is 176.46  so a at 0°=1127.392Nm/176.46 = 6.389 I don't know what the units are here! deg/s^2? rad/s^2? m/s^2 a at 30°=972.33Nm/176.46 = 5.51 a at 60°=548.696Nm/176.46 = 3.11 a at 90°=(30Nm)/176.46 = 0.17 and from these accelerations (if I even did this part correctly), I figured I could plug them into the equation a=Δw/Δt so w=aΔt (is it okay to disregard the Δ "change in" here? or no?) w at 0°=6.389*0.1 = 0.6389 **units?** w at 30°=5.51*0.1 = 0.551 is it okay to use t=0.1s or do I use t=0.2s? w at 60°=3.11*0.1 = 0.311 w at 90°=0.17*0.1 = 0.017 I see in my notes that the units are supposed to be in rad/s, but I'm not sure what units I even have right here, so I can't even begin to convert these numbers if I needed to.. please let me know if I did anything right/wrong.. I am feeling very lost here.. thank you! 



#4
Nov2311, 09:49 PM

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P: 11,416

Angular Velocity and Torque homework question
The normal units of angular acceleration are rad/sec^{2}. A torque in N m divided by a moment of inertia in kg m^{2} will give an acceleration [itex]\alpha[/itex] in radians/sec^{2}. Degrees/sec^{2} are acceptable if the problem is using degrees for its angle measurements, but you need to be a bit careful about when they 'pop out' naturally in an equation and when you'll need to apply a conversion. You should be sure to make clear what units you use.
I think your moment of inertia should be a tad higher than the 176.46 kg m^{2} you've stated. I see it being closer to 180 kg m^{2}. The equations of motion in the angular domain are a direct parallel to those in the linear domain. So if you have a moment of inertia (mass) and a force (torque), then you you can determine the angular acceleration (acceleration). a = f/m > [itex]\alpha = \tau/I[/itex] . Similarly, given an acceleration and a time interval, in the linear motion domain you would write [itex]v = v_i + a \Delta t[/itex], while in the angular domain you would write [itex] \omega = \omega_i + \alpha \Delta t [/itex]. Part 2 of the problem, as you describe it, is a bit odd. Still, is is what it is I would point out that for the initial position (when the body is initially horizontal) that there is no 'previous position' from which a torque, and thus an acceleration, can be obtained. But for the rest of the positions you can certainly calculate the 'legacy' torques and angular velocities. Given an initial angular velocity and torque, then [itex] \omega = \omega_o + \alpha t[/itex]. So this is the equation you should be using to find the 'current' angular velocity after the last position, given that the specified torque is applied for 0.1 seconds. The angular velocities should be cumulative. In other words, you need to apply each velocity change to the existing velocity as you go. 



#5
Nov2311, 10:11 PM

P: 5

so you are saying that at 0° from horizontal, I should have an angular acceleration and an angular velocity of 0?
..and then for 30° from horizontal, when I use the equation ω=ω_{o} + αt, the ω_{o} would = 0? .. and sorry, the time I use would always be 0.1s, or do I need to add 0.1s every time I calculate the angular velocity? my moment of inertia might be a bit off because he told use to use a radius of gyration that is 10cm further from our hands than our COM.. so I used the formula I=mk^2 to find this. When I did all of my cos calculations, my calculator was in degrees, so I'm assuming that my accelerations are in deg/s^2.. and so I should convert all of my accelerations/velocities to rad/s^2 and rad/s.. 



#6
Nov2311, 10:33 PM

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P: 11,416

When you calculate an acceleration by dividing the torque by the moment of inertia, the result will be in radians. Radians are the "natural" units for angles that result from such calculations. You'd have to convert these results to degrees if you want degrees. It's generally easier to just leave everything in radians until a final result is to be presented using degrees as the angular unit. 



#7
Nov2411, 12:36 AM

P: 5

okay, thank you so much for your help..
and I just realized, I forgot that I rounded my radius of gyration in this forum to 1.4m.. it is really 1.388 in my notes, and that calculation works out. also, since α=T/I, I am finding an angular acceleration of 6.388rad/s^2 at the 0° from horizontal point.. how is it possible to have acceleration when there is no velocity? so when I calculate the angular velocity for each level, for example if I am calculating it at 30° from horizontal, would I use the α from 30° or the α from 0°? (angular accel. at 0° = 6.388rad/s^2) (angular accel. at 30° = 5.51rad/s^2) ω at 30° = 0 + (6.388rad/s^2 * 0.1s) or ω at 30° = 0 + (5.51rad/s^2 * 0.1s) one more thing: the question says to calculate torque every 30° until the rotation "stops or changes direction" and I know that a counterclockwise rotation is considered positive, and a clockwise rotation is considered negative.. so at 90°, when the torque = 30Nm, can I say that this is the point where rotation changes direction? ...though, surely the gymnast doesn't actually start rotating backwards before she reaches the bottom.... 



#8
Nov2411, 06:21 AM

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#9
Nov2411, 10:36 PM

P: 5

okay I got all of the torques and velocities for each different angle, and it turned out that I had not calculated enough angles initially.. I ended up going all the way up to 240° where the velocity became negative, so I knew that a change in direction had occurred.
there is one more part to this question, and it goes like this: "recalculate the ascending phase of the swing but assume you take a pike position (flexion at the hips, bring the legs up) instantaneously at 90° (vertical), and assume that your center of mass is 20cm closer to the hands. Also, assume that your radius of gyration is 10cm further from the hands than the center of mass." so, without doing calculations here, I'm assuming that my angular velocity should increase, and my moment of inertia will decrease. The pike will decrease the moment arm, and the torque values will also decrease.. when I used my torque values to find angular acceleration, then used those values to calculate the angular velocity, my velocities were actually decreased compared to when i originally didn't pike in the ascending phase. I don't know what I'm doing wrong here... 



#10
Nov2511, 08:01 AM

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P: 11,416

There should be a sudden increase in his velocity when he "instantaneously" assumes the pike position. Use conservation of angular momentum to determine how the velocity changes at the 90° position.




#11
Jun513, 08:41 PM

P: 2

I have a question about the first half of the question, what are you talking about with a triangle i am completely lost with that. I have the torques for the one side but do i have to do the torques for when it goes it the other side cause it states when it changes direction or stops but it is still going in that direction so would i have to find the torque also?




#12
Jun513, 08:43 PM

P: 2

I say that because in this question it states that the gymnast starts from 90 degree from vertical




#13
Jun513, 09:11 PM

HW Helper
P: 6,925




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