posted by Anubhav .
A 15 kg, 1m wide door which has frictionless hinges is closed but unlocked. A 400 g ball hits the exact middle of the door at a velocity of 35 m/s and bounces off elastically, thereby causing the door to slowly swing open. How long in seconds does it take for the door to fully open (rotate 90 degrees)?
The force exerted by the hinges has a torque of zero relative at the hinges. This means that angular momentum is conserved w.r.t. all point on the line between the two hinges. We also have conservation of energy. The moment of inertia of the door w.r.t. the rotation axis is:
I = 1/3 m L^2
where M = 15 kg is the mass of the door and L = 1 m is the width of the door.
Let's denoting the initial velocity of the ball in the forward direction (the direction where the door opens) by Vi, the velocity after it bpuncess off the door in the opposite direction as Vf. Let's put Mb = 400 g = mass of the ball and we put R = 1/2 m = radius at which the ball hits the door.
We then have:
Conservation of angular momentum:
Mb Vi R = I omega - Mb Vf R (1)
Conservation of energy:
1/2 Mb Vi^2 =
1/2 I omega^2 + 1/2 Mb Vf^2 (2)
From (1) it follows that:
Vf = I omega/(Mb R) - Vi
Inserting in (2) gives:
1/2 I Omega^2 +
1/2 Mb (I omega/(Mb R) - Vi)^2 =
1/2 Mb Vi^2
If you expand this out, you can divide bvy Omega (omega = 0 is an unphysical solution corresponding to the ball moving through the door without interacting with it). You then get:
Omega = 2 Vi/[R(1 + I/(Mb R^2))]
The time to rotate through 90 degrees is pi/(2 Omega).
hey,@ count ibis, don't u think your omega answers value is too big to open the door slowly,
what is answer
wrong.....i answer 0.5722 too?
@- Count Iblis do you know the answer?
I got the distance the door travels is pi/2 x 1m which is the radius since the door will travel in a circle, then since the momentum is conserved, the velocity of the door will be 14/15, and then pi/2 divided by 14/15 gives 1.68
but 1.68 is wrong answer
Finally 0.572 is correct