- #1
ac7597
- 126
- 6
- Homework Statement
- Bob creates a square frame by connecting four metersticks, each of mass m=0.1 kg, at their ends. He drills a hole at one corner, then suspends the square by a pin through that corner. The square hangs down motionless, so it looks like a diamond.
What is the moment of inertia of the square frame around the pin?
How far below the pin is the center of mass of the square?
Bob pulls the bottom of the frame to the side a short distance, then releases it. The square frame oscillates back and forth, back and forth. What is the period of the motion?
- Relevant Equations
- angular velocity = (m*g*d/inertia)^1/2 = 2pi/period
inertia of center = [(1/12) m*L^2 + m(L/2)^2]*4
inertia of center = (4m*L^2)/ 3
inertia around pin = (4m*L^2)/ 3 + 4m(L/ 2^(1/2) )^2
inertia around pin = (10m*L^2)/ 3
inertia around pin = (10*0.1*1^2)/ 3 = 0.33 kg*m^2
d= 1/2^(1/2) = 0.707m
(m*g*d/inertia)^1/2 = 2pi/period
(0.1*9.8*0.707/0.33)^1/2 = 2pi/period
period=4.35 seconds
Apparently 4.35s is not the right answer. I don't know why.
inertia of center = (4m*L^2)/ 3
inertia around pin = (4m*L^2)/ 3 + 4m(L/ 2^(1/2) )^2
inertia around pin = (10m*L^2)/ 3
inertia around pin = (10*0.1*1^2)/ 3 = 0.33 kg*m^2
d= 1/2^(1/2) = 0.707m
(m*g*d/inertia)^1/2 = 2pi/period
(0.1*9.8*0.707/0.33)^1/2 = 2pi/period
period=4.35 seconds
Apparently 4.35s is not the right answer. I don't know why.