- #1

linds43

- 5

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## Homework Statement

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

## 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!

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