In the see-saw shown below the mass of the metal bar is 6.0 kg and the other two
ID: 1602351 • Letter: I
Question
In the see-saw shown below the mass of the metal bar is 6.0 kg and the other two masses are m1 = 4.0 kg and m2 = 2.0 kg. The see-saw is released from rest and there is no friction at the point of rotation.
Showing your thinking clearly is required for all questions. 1. In the see-saw shown below the mass of the metal bar is 6.0 kg and the other two masses are m 4.0 kg and m2 2.0 kg. The see-saw is released from rest and there is no friction at the point of rotation. 1.0 m a. What is the direction of the net torque acting on the see-saw? b. You will now calculate the angular speed of the see-saw when it becomes vertical with mi at the bottom. i. Choose the zero of gravitational potential energy to be the lowest point reached by mi. Write out an expression for the initial gravitational potential energy and another for the final gravitational potential energy. ii. Write out an expression for the final kinetic energy. Note that mi and m2 must be moving at the same speed since they are at the same distance from the pivot. Also, use the fact that v where L is the length of the rod to simplify your expression iii. What is the linear speed of each mass in the final situation?Explanation / Answer
(A) m1 is heavier hence torque is counter clockwise direction.
(B) initial PE = m1 g L/2 + m2 g L/2
= g L/2 ( m1 + m2)
final PE = m2 g L
(ii) Applying energy conservation,
PEi + KEi = PEf + KEf
g L /2 (m1 + m2) + 0 = m g L + KEf
KEf = g L/2 (m1 - m2)
(iii) KEf = m1 v^2 /2 + m v2^2 / 2 + I w^2 /2
g L/2 (m1 - m2) = m1 v^2 / 2 + m2 v2^2 /2 + ( M L^2 / 12) (v / L)^2 / 2
9.8 x 0.5 x 2 = 3 v^2 + v^2 + 0.25 v^2
v =1.52 m/s
C. if we increase M then v will decrease/