physics question

Question

A 3.91-m long thin seesaw with a mass of 44.9 kg has a pivot in its center. Two children with masses 38.9 kg and 30.7 kg sit on extreme opposite ends and start moving up and down. When the seesaw is horizontal and neither child is in contact with the ground, what is the magnitude of the net torque acting on the board (use units of N*m)? What is the magnitude of the angular acceleration of the seesaw at this instant (use units of rad/s^2)?

Answer 1

Assign rotation directions randomly because questions are about magnitudes so direction does not matter.

torque clockwise (torcw) = weight x lever arm = 30.7(9.8)(3.91/2) = 588.2 Nm

torque counterclockwise (torccw) = weight x lever arm = 38.9(9.8)(3.91/2) = 755.3 Nm

net torque (tornet) = 755.3 - 588.2 = 157.1 Nm

tornet = I∝ so to find ∝, we need to find moment of inertia (I)

Because the axis of rotation of the seesaw is also its center of mass, we can ignore its mass

I = mr² + mr² = (30.7)(3.91/2)² + (38.9)(3.91/2)² = 277 kgm²

∝ = tornet/I = 157.1/277 = 0.57 rad/s²