Question:

A thin rod of length L and mass M is initially at rest. It is then free to rotate about one end, where a force F is applied perpendicular to the rod at a distance d from the end. The rod is initially at an angle θ with respect to the horizontal. Find the angular acceleration α of the rod about its pivot point in terms of M, L, F, d, and θ.

Answer:

To start, we can identify the torque on the rod about its pivot due to the applied force F. The torque τ is given by the cross product of the position vector r from the pivot to the point where the force is applied and the force vector F:

τ = r × F

In this case, the position vector r has magnitude d, and the direction points perpendicular to both r and F, which in this case is into the page due to the rotation of the rod. Therefore, the torque is given by:

τ = rF

Next, we can apply Newton's second law for rotational motion, which states that the net torque τ about the pivot point is equal to the moment of inertia I of the rod times its angular acceleration α:

τ = Iα

The moment of inertia I of a thin rod rotating about one end is given by I = (1/3)ML^2.

Equating the torque from the applied force to the moment of inertia times the angular acceleration, we have:

rF = (1/3)ML^2α

Now, the position vector r in terms of θ can be expressed as r = Lsin(θ). Substituting this into the equation, we have:

Lsin(θ)F = (1/3)ML^2α

Finally, solving for the angular acceleration α, we get:

α = (3Fsin(θ))/(ML)

Thus, the angular acceleration α of the rod about its pivot point is given by α = (3Fsin(θ))/(ML).