Can anyone help me with this problem?

Question

Consider the forces on a banging door. Apart from gravity, there is a force from the surface that the door strikes and another force on the hinges on which the door swings. Let the door have width d, mass M and moment of inertia I. The stopper is placed at the half-height of the door but at an unknown radial distance d from the hinge. Suppose that just before striking the door stopper, the door has angular velocity w0. The forces exerted by the hinges and stopper that bring the door to rest will be denoted Fh and Fs. Before the door stops, there is also a small radial force Fr, exerted by the hinges, but we’ll ignore it in our calculations. We’ll also ignore the hinge forces that keep the door from tilting in its own plane. (b) Now consider the linear momentum of the door. Write down an equation which relates the linear momentum of the door’s center of mass (at radial distance w/2 from its hinges) to its mass and rotation frequency. That linear momentum is directed up the page.

Answer 1

Hi Yogita S.,

So what's the issue here? You know the mass of the door, and you know that the center of mass is moving in a circle of radius (d/2) at angular frequency (w(0)). That's 1 total circle (2.pi.r) in one period. Calculate the instantaneous (linear) speed of that (CM) spot, times the mass. (it's "up the page")

The fun comes when you get to Fh and Fs. Until you get more data, these aren't forces, they're impulses, since they have to counteract angular momentum via exerting torque. If the frame and the stopper don't have some "give"=elasticity, the door rips off the hinges, and/or punctures/disassembles at the stopper target point.

Just sayin'. -- Cheers, -- Mr. d.