After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod 0.83 m long that pivots freely about the hip.
KE = 0.5mv^2(1 + I/(mr^2))
PE = mgh
I = (1/3)mr^2
The Attempt at a Solution
This problem is driving me crazy! I've tried so many different ways, and I can't seem to get it. Since the problem gives us that it's a uniform rod that pivots around the hip, I equated that to the rod being rotated on an axis at an endpoint. Hence me getting I = (1/3)mr^2.
From there, I basically plugged and chugged. Since it's starting at a height, the initial PE = mgh, and initial KE = 0. The final KE = the long equation in 2, and the final PE = 0 since it lands on the floor.
initial PE + initial KE = final PE + final KE
mgh + 0 = 0 + 0.5mv^2(1 + I/(mr^2))
mgh = 0.5mv^2(1 + I/(mr^2))
The two masses then cancelled out to give the following:
gh = 0.5v^2(1 + I/(mr^2))
Since I = (constant)mr^2, the two terms cancel out, so all that's left is:
gh = 0.5v^2(1 + (constant))
(9.81)(0.83) = 0.5v^2(1 + (1/3))
v = [2(9.81)(0.83)/(4/3)]^0.5
And when I solved for v, I got 3.49 m/s. But this isn't the right answer. I'm not sure why it isn't, and I'm not too sure what to do now. I thought maybe that the leg was only doing rotational kinetic energy since it wasn't really doing any linear movement, but that sort of got me confused, so I abandoned that trail of thought.
Some help would be very appreciated. Thanks!