Rotational Dynamics

Question

Ann and Bob are pushing opposite sides of a door with I=10.5 kg•m^2. Ann pushes 34.8N at 0.732m from the axis, while Bob pushes 59.6N at 0.300m, both perpendicular to the door. What is the angular acceleration? [?] rad/s^2

Hint: Remember, CCW torque is +, CW torque is -.

Answer 1

The main equation to use in the problem is τ=Iα where τ is torque, I is moment of inertia (given in the problem) and α is the angular acceleration (which we are looking for). Our first step is to solve for the net torque.

τ = rFsinθ

Since both are pushing the door perpendicularly, θ=90° and sin(90°)=1, therefore our torque equation simplifies to:

τ=rF

Since we have two forces, we have 2 torques and our net torque will be the sum of the 2.

To determine the signs: Imagine looking down at the door from the top where the hinge is at the top of the page. If Ann is on the lefthand side of the door pushing towards the right this would cause the door to swing counterclockwise. Bob is on the opposite side pushing towards the left causing the door to swing clockwise. This means that Ann's torque is positive and Bob's torque is negative (see the hint at the bottom of the problem).

Net torque can then be calculated as:

τ_net = +r_Ann(F_Ann) - r_Bob(F_Bob) = 0.732(34.8) - 0.300(59.6) = +7.59 Nm

Then use the initial formula of:

τ=Iα

7.59 = 10.5α

so α=0.722 rad/s² in the CCW direction (since our answer is positive)

Hope this helps!