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...
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...
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=ωo + αt, the ωo would = 0? .. and sorry, the time I use would always be 0.1s, or do I need to add 0.1s every time I...
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...
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...