Kinematics question -- Impulse applied to a plate hanging from a wall

[mcchoy528]

Homework Statement

A circular plate with radius 0.5 m and mas 5 kg is hung on the wall, fixed at a point that is 0.3 m above its center. The plate can freely rotate about the fixed point with no friction. A very short-duration impulse of 5Ns, along a direction that is tangential to the circumference of the circular plate, is applied at the bottom point of the plate. From energy conservation, what is the maximum angle of rotation (away from the equilibrium position) attained by the plate?

Homework Equations

J=mΔv

The Attempt at a Solution

I_p=1/2 0.5²*5+5*0.3²=1.075
1/2 mv²=1/2 Iω²
ω=2.16rad/s
How can I calculate the duration of the impulse?

 

[VKint]

You shouldn't need to know the duration of the impulse.

Instead, think about the information provided that you haven't used yet. For example, why do you think it's relevant that the impulse is tangential to the circular plate? Also, the problem suggests using conservation of energy, but impulse is a concept more closely related to momentum. What is the relationship between kinetic energy and linear momentum? How about between kinetic energy and angular momentum?

 

[haruspex]

ω=2.16rad/s

How did you deduce this?

 

[Delta2]

The rotational kinetic energy is $K_{rot}=\frac{L^2}{2I}$ where $L=I\omega$ is the angular momentum of the plate immediately after the impulse acts.
The translational kinetic energy is $K_t=0$ (we might have to talk in more detail about this) .

From conservation of energy we can see that all the kinetic energy will become gravitational potential energy, so by $K_{rot}+K_t=U=mgh$ we can find the new height $h$ of the center of mass (relative to the center where we take the potential energy to be zero), and then relate $h=h(\theta)$ to the maximum angle $\theta$ attained and solve for the angle $\theta$.

All that left to be done is to carefully relate $L$ to the impulse 5Ns. Any ideas you got on that?