A 2.5-m long board with a mass of 12 kg is placed on a bar so that it is perfect

Question

A 2.5-m long board with a mass of 12 kg is placed on a bar so that it is perfectly balanced, i.e., the bar exactly underearth the middle of the plank:

a. A 22-kg child sits on the left end of the board. What is the torque they exert on the system?

b. A second child with a mass of 27 kg sits on the other side. How far away from the center must the second child be if the board is to remain in equilibrium?

c. Suppose the second child sits on the opposites end of the board. What is he net torque acting on the system?

d. What is the moment of inertia of the board? If we each child as a point mass, what is the total moment of inertia of the system?

Explanation / Answer

a) Torque is 22* 9.8*(2.5/2) = 269.5 Nm

b) the second child must sit at distance of =269.5/ (27*9.8) = 1.018 m from the centre

c) net torque is = (27- 22)*(2.5/2) = 6.25N m

d)moment of inertia of the bar is -= m*l*l /12 = 6.25 kgm^2

the total moment of inertia of the system = MOI of bar + MOI of child = 6.25 + 22*(1.25)^2 + 27*(1.25)^2 = 82.81 kgm^2

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