Year1 Mechanics (angular momentum)

[AlanWWW]

Homework Statement

A circular plate with radius 0.5 m and mass 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 5 N sec, 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?

r = 0.5 m
m = 5 kg
Impulse = 5 Ns

Homework Equations

w is omega
r is the radius

Moment of Inertia = I_cm + MR^2
Rotational energy = 0.5*I*w^2
angular momentum = linear momentum*r
angular momentum = I*w (if symmetry)

The Attempt at a Solution

[/B]
First, find the moment of inertia by I = I_cm + MR^2

I = 0.5*m*r^2 + m*r^2
= (0.5)(5)(0.5^2)+(5)(0.3^2)
= 1.075Then, find the initial angular velocity

By impulse = Δangular momentum
(I stuck here because the equation" angular momentum = I*w " cannot be used
as it only works when the rotation axis is symmetry)Then the question said from conservation of energy, so it should relate to:
Initial Energy = Final Energy

Initial Energy = E_rotational + E_transitional
(Energy of transitional should not be zero as the center of mass is moving,
I am not sure it is correct or not.)

Final Energy = m*g*h
(However I don't how this equation apply to a rigid body but not a point mass)

 

[kuruman]

By impulse = Δangular momentum
(I stuck here because the equation" angular momentum = I*w " cannot be used
as it only works when the rotation axis is symmetry)

Yes it can. Angular momentum needs an axis for its definition and so does the moment of inertia. As long as both are defined with respect to the same axis, the equation is valid.

Final Energy = m*g*h
(However I don't how this equation apply to a rigid body but not a point mass)

What is h? The plate has many points. Hint: At what point does the external force of gravity act?

A very short-duration impulse of 5 N sec

The duration of the impulse is irrelevant if you are not given the force that acts during the 5 ns interval. If you hit the plate with all your might as opposed to lightly tapping it for 5 ns, the results will be different, no?

 

[AlanWWW]

Yes it can. Angular momentum needs an axis for its definition and so does the moment of inertia. As long as both are defined with respect to the same axis, the equation is valid.

What is h? The plate has many points. Hint: At what point does the external force of gravity act?

The duration of the impulse is irrelevant if you are not given the force that acts during the 5 ns interval. If you hit the plate with all your might as opposed to lightly tapping it for 5 ns, the results will be different, no?

h is the vertical displacement of center of mass

The question gave impulse only, so I think the impulse-momentum theorem should be used

 

[kuruman]

Sorry, I misinterpreted 5 N sec as 5 nanoseconds. I see now that it's 5 Newtons⋅seconds. Yes, use the impulse-momentum theorem. Can you solve the problem now?

 

[AlanWWW]

No... there is problem about the value of h
I calculate with following steps:

I = 0.5*m*(r^2) + m*(r^2) <---first r is 0.5 and second r is 0.3
I = 1.075

I*w = impulse
w = 4.651

KE_initial = KE_final
0.5*I*(w^2) + 0.5*m*(v^2) = mgh
0.5*I*(w^2) + 0.5*m*(r^2)*(w^2) = mgh <--r is 0.3
h = 0.34
which is larger than 0.3 , that's means the center of mass raise above the axis of rotation
so the angle of rotation is larger than 90 degree
However, the solution of this question is C(60.4 degree)

 

[kuruman]

I*w = impulse
w = 4.651

There is a problem here. Iω has unit of angular momentum while impulse has units of momentum. You cannot set them equal. For linear motion you have linear impulse = change in linear momentum. For angular motion you have angular impulse = change in angular momentum. Any ideas how to find the angular impulse from the linear impulse? Hint: How do you find angular momentum from linear momentum?

 

[AlanWWW]

There is a problem here. Iω has unit of angular momentum while impulse has units of momentum. You cannot set them equal. For linear motion you have linear impulse = change in linear momentum. For angular motion you have angular impulse = change in angular momentum. Any ideas how to find the angular impulse from the linear impulse? Hint: How do you find angular momentum from linear momentum?

The angular impulse = radius of rotation *linear impulse

 

[AlanWWW]

After the adjustment , i got h
Then I calculate the angle:

arcosθ=(0.3-h)/0.3
And I got θ = 73.6 degree, which is wrong

However
If I do not calculate the transitional kinetic energy:
KE_rotational = mgh
I got the exact correct answer
Is there any conceptual mistake I made?

 

[kuruman]

Is there any conceptual mistake I made?

There is no mention that kinetic energy is conserved during the collision, so you cannot assume that it is. However, you may assume that angular momentum is conserved because the impulse is of very short duration, so short that the external torque due to gravity does not affect the angular velocity as it is being acquired by the plate.

 

[AlanWWW]

Thx