Angular speed (Physics problem)?
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well.
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.4 {\rm kg\cdot m^2/s} and for arms and legs in is 0.70 {\rm kg\cdot m^2/s}. If she starts out spinning at 4.7 {\rm rev/s}, what is her angular speed (in {\rm rev/s}) when her arms and one leg open outward?
Comments
you forgot to factor in vector coefficient*(2a+l) where a=arm and l=leg, also, isn't a factor of 4.7 an impossible revolutionary speed? dizziness would cause enough rotational instability that there wouldn't be a constant factor that could be relied on. in addition, the extension length of the arms and legs will be another factor, as well as the weight of the skater and weight of arms and leg, these also have a huge variable in terms of rotating mass. try revamping your hypotheticals to a more realistic set of data along with the values of extremity mass and i will try to resolve your equation for you.
till now the 1st start up, we could desire to stipulate our very own course, so i define the clockwise as adverse and counter-clockwise could be my valuable course. Disk A, m=8.0kg, R=20cm, w= -30rev/s = -60pie/s Disk B, m=8.0kg, r=10cm, p=30rev/s = 60pie/s via utilising the thought of conservation of momentum, the preliminary and the magnificent momentum is the comparable, so u have been given this equation, (8)(0,2 x -60pie/s) + (8)(0.a million x 60pie/s) = (8 + 8)V V= -3pie / s V= -3rev/s rotate interior the clockwise course...XD