Question

A force F is applied to the end of a thin rod of length L and mass M that is lying
on a frictionless surface. What is the angular acceleration of the rod? At the instant shown, what is the linear acceleration of the end of the rod
opposite the point where the force is applied?

Answer 1

Using laws of dynamics:
First, write torque about centre of mass of the rod, and find angular acceleration \(\alpha\).
Second, find acceleration of centre of mass \(a\).

Using kinematics:
Find acceleration of opposite end of the rod by combining \(a\) and \(\alpha\) above.

Answer 2

Radius of Gyration(for rotation about the centre of mass) of the rod(assuming homogeneous density ) = \[L^2/12\]
torque about the centre of mass = \[\Gamma = F * L/2\]
torque would produce angular acceleration.
\[\Gamma = I * \alpha\]
\[I = M*L^2/12\]
hence,
\[\alpha = 6*F/(M*L)\]

Now, linear acceleration of the centre of mass = F/M

for the opposite end of the rod, acceleration of the rod with respect to the centre of mass =
\[-L/2*\alpha\]

so, at this instant, total acceleration of the opposite end of the rod = acceleration relative to centre of mass + acceleration of centre of mass at
= -6F/2M + F/M
\[acceleration = -2*F/M\]