Rotational Dynamics

We begin be calculating the magnitude of each torque individually. Recall that τ = Fr sinθ . Thus τ ₁ = (30)(1)sin 120 = 26.0 N-m and τ ₂ = (50)(1)sin 30 = 25 N-m. As we can see from the figure, τ ₁ acts counterclockwise while τ ₂ acts clockwise. Thus the two torques act in opposite directions, and the net torque is thus 1 N-m in the counterclockwise direction.

Since both cylinders have the same shape, they will experience the same forces, and thus the same net torque. Recall that τ = Iα . Thus the cylinder with the smaller moment of inertia will accelerate more quickly down the incline. Think of each cylinder as a collection of particles. The average radius of a particle in the solid cylinder is smaller than the hollow one, as most of the mass of the hollow one is concentrated at a larger radius. Since moment of inertia varies with r ² , it is clear that the solid cylinder will have a smaller moment of inertia, and thus a larger angular acceleration. The solid cylinder will reach the bottom of the incline first.

We begin by resolving the gravitational force into tangential and radial components, as shown below:

Recall that only the tangential component of a force will produce a torque. The magnitude of the tangential component is given by F sinθ = mg sinθ . This force acts at a distance r from the axis of the rotation. Thus the magnitude of the torque is given by:

We already know the torque acting on the pendulum. Recall that τ = Iα . Thus, to find the angular acceleration we need to compute the moment of inertia of the pendulum. Fortunately, it is simple in this case. We can treat the mass on the pendulum as a single particle of mass m and radius r . Thus I = mr ² . With this information we can solve for α :

Although it looks like the forces are directed in opposite directions, and thus cancel out, we must remember that we are working with angular motion here. In fact, both forces point in the counterclockwise direction, and can be considered to have the same magnitude and direction. In addition, they are both perpendicular to the radial direction of the door, so the magnitude of the torque by each one is given by: τ = Fr = (50 N)(1 m) = 50 N-m. As we stated, the two forces act in the same direction, so the net torque is simply: τ = 100 N-m.

Next we have to calculate angular acceleration. We already know the net torque and thus must find the moment of inertia. We are given the formula I = . We are given the mass, and from the figure we see that the radius is simply 1.5 m. Thus: