1. zero is correct
2. yes, any force up from the floor has no moment arm
3. change in Pe = m g h = (1/2) m g L
Ke = (1/2) m v^2 + (1/2) I w^2
I = (1/12) m L^2
v = (L/2)w
so
Ke = (1/2)m(L^2/4)w^2 + (1/2)(1/12)m L^2 w^2
= (m L^2 w^2) (1/8 + 1/24)
= m L^2 w^2 (1/6)
so
(1/2) m g L = m L^2 w^2 /6
w^2 = 3 g /L
w = sqrt (3 g/L)
A thin rod (uniform density & thickness) has mass M and length L. It is attached to the floor at a fixed location by a friction-less hinge. The rod start at rest, balanced vertically on its hinge end.
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1) While balanced vertically, the gravitational torque acting on the rod is: MY ANSWER - I said that it had 0 (zero) torque. Is this correct?
Now, the rod is given a nudge to the right, so that it starts to fall with nearly zero initial speed. Ignore air resistance.
2) Just before the rod hits the floor, the gravitational torque acting on the rod is: MY ANSWER - I said that the torque was 1/2MgL. Is this correct or is the coefficient wrong?
3) What is the final angular velocity of the rod just before it hits the floor? Express the final answer in terms of variable L, M, g, and mathematical constants. (Hint: Use conservation of energy)
MY ATTEMPT - Conservation of energy is expressed by: (U_grav)i = (K_trans+K_rot)f ... where do I go from there?
1 answer