How Much Energy Is Lost When a Sledgehammer Hits a Stationary Merry-Go-Round?

[Tanero]

Homework Statement

A merry-go-round is sitting in a playground. It is free to rotate, but is currently stationary. You can model it as a uniform disk of mass 220 kg and radius 100 cm (consider the metal poles to have a negligible mass compared to the merry-go-round). The poles near the edge are 86 cm from the center.

Someone hits one of the poles with a 9 kg sledgehammer moving at 15 m/s in a direction tangent to the edge of the merry-go-round. The hammer is not moving after it hits the merry-go-round.

How much energy is lost in this collision? (enter a positive number for the absolute value in Joules)

Homework Equations

conservation of momentum
KE of rotation =.5mv²
KE of linear motion = .5Iω²
Lost of energy=KE_rot-KE_linear

The Attempt at a Solution

I assume that hummer hit pole tangentially and collision was elastic.
My way of thinking is that energy transfer of KE of linear motion to KE of rotation. Thus I need to find KE_linear and KE_rot. Where KE_linear is easy to find.

To find KE_rot, first, apply conservation of momentum.
m_hv_h=m_dv_d
Linear momentum was transferred to rotational motion. Where v_d is tangential velocity of disk. v_d is velocity of hummer. Then, v=ωd (d-distance from center to metal pole)
From these, I can find ω and put it into my KE_rot

Than I can find my energy loss. Right?

 

[jedishrfu]

Minor nit but you wrote your relevant equations a bit off interchanging linear KE and rotational KE equations

From what the problem said all the linear KE is transferred to the rotational KE so that means

KE linear = KE rotation

so you need to compute the I in the Ke rotation knowing that its a disk of a certain mass and diameter rotating about a certain axis.

 

[Tanero]

@ jedishrfu
I don't think these KE are equal.

I found, that KE of rot is less than KE linear. Maybe because the force is applied not at the rim but at some distance. That's why I got some energy loss.

Is my approach to find omega right? omega is only thing missing in KE of rot.

 

[mfb]

I moved your thread to introductory physics homework.

conservation of momentum

Momentum can be transferred to the axis of rotation, it does not have to be conserved.

I assume that hummer hit pole tangentially and collision was elastic.

If the collision would be elastic, no energy would be lost.

From what the problem said all the linear KE is transferred to the rotational KE so that means

KE linear = KE rotation

That approach does not work. The problem statement directly says that energy is lost (and you can verify this with a calculation).Angular momentum is the correct approach.

 

[haruspex]

Momentum can be transferred to the axis of rotation, it does not have to be conserved.

Just to clarify mfb's remark...
When the hammer struck the pole, the merry-go-round did not go sliding across the park. This is because there was an equal and opposite (parallel) reaction from the axle in the middle. That's why we can't use conservation of linear momentum here - we don't know the magnitude of that impulse. We can eliminate it from consideration by taking moments about the axle. It has no moment about that point.

 

[Tanero]

@ haruspex,
if I consider torques about center of rotation, I then come to conclusion that angular momentum right before the collision equals angular momentum right after, right?

 

[mfb]

Torques don't help (you do not have a collision time), but angular momentum before and after the collision is the same, right (if you define your axis properly).

 

[Tanero]

@mfb
I considered that angular momenta before and after are conserved about center of rotation of the merry-go-round. I don't see any other axis helpful. What do you mean properly?

 

[haruspex]

@mfb
I considered that angular momenta before and after are conserved about center of rotation of the merry-go-round. I don't see any other axis helpful. What do you mean properly?

I believe that's the axis mfb had in mind. Perhaps "smartly" would have been clearer.
But mfb's point is that the hammer supplied an impulsive angular momentum, i.e. a change to the angular momentum in an unknown short time. Hence the torque, as such, cannot be known, but is uninteresting.
So, what is the total angular momentum of the system (merry-go-round+hammer) about the axis just before impact?

 

[Tanero]

Angular momentum of the system is just product of distance to the axis and moment of hammer before the impact.

 

[haruspex]

Angular momentum of the system is just product of distance to the axis and moment of hammer before the impact.

Yes.