Bike Wheel Question: Kinetic Energy Lost on Impact

[rtsswmdktbmhw]

Homework Statement

Bicycle wheel is at rest, and can rotate frictionlessly about a fixed axle. A dart travels at velocity v (in plane of wheel) parallel to a line that goes through the centre of the wheel. It hits the rim of the wheel and sticks.

What kinetic energy, if any, is lost during impact?

Homework Equations

The Attempt at a Solution

I thought about something like 'the kinetic energy whose component is perpendicular to the rim of the wheel at impact is lost', but energy is scalar so that didn't seem right. Is there some better worded solution/reasoning? Perhaps the fixed axle buffers the wheel and so some kinetic energy is lost?

Or, am I supposed to answer these types of questions simply by directly calculating before/after?

Also (as an aside), is it correct to take the angular momentum of the dart simply to be about the axis of the wheel, and = $\vec r\times \vec p$ at the instant it hits the rim? (I haven't included the relevant question here.)

 

[andrevdh]

The dart comes to standstill and the wheel is not rotating! Read! Think! Don't just reproduce info.

 

[NascentOxygen]

The dart comes to standstill and the wheel is not rotating! Read! Think! Don't just reproduce info.

The dart is not^✶ brought to a complete standstill, it starts the wheel turning.

 

[andrevdh]

That is what happens if you do not read carefully!
Some energy will be lost due to the work being done against friction upon entry,
but it will be minimal due to the sharp point.

 

[NascentOxygen]

I think the expectation is for some equations to quantify the loss of energy. Perhaps the loss could be related to the perpendicular distance of the incoming dart's path from a parallel path through the wheel axle.

 

[rtsswmdktbmhw]

That is what happens if you do not read carefully!
Some energy will be lost due to the work being done against friction upon entry,
but it will be minimal due to the sharp point.

What friction is this? Is it due to the fact that the dart stuck to the rim? Wikipedia says that inelastic collision is one where the two objects e.g. carts stick after collision, but I wasn't sure if that would apply here since that dart has a sharp end.

I think the expectation is for some equations to quantify the loss of energy. Perhaps the loss could be related to the perpendicular distance of the incoming dart's path from a parallel path through the wheel axle.

Yes, this is usually what I would do as it's probably the most straightforward way.. But some of the solutions I have been given to similar questions have been qualitative, so I wasn't too sure if quantifying it was the expected approach.

If quantifying, then it would just be 0.5*m*v^2 of dart = 0.5*I*omega^2 (wheel + dart after collision) right?

 

[NascentOxygen]

If quantifying, then it would just be 0.5*m*v^2 of dart = 0.5*I*omega^2 (wheel + dart after collision) right?

Energies? You're assuming energy is conserved, but it won't be. The whole aim of the exercise, IMO, is to arrive at a formula that shows what percent of the energy is lost. (Remember, a "currency" that is conserved during the exchange is momentum.)

If the dart were incoming on a path directed through the wheel's axis, then all energy would be lost. The dart would halt and the wheel would not be turned.

 

[rtsswmdktbmhw]

Oh, whoops, I meant comparing, not equating, the 0.5*m*v^2 (before collision) with 0.5*I*omega^2 (after collision), where I is the rotational inertia of wheel+dart and omega can be worked out via conservation of angular momentum.

 

[haruspex]

That is what happens if you do not read carefully!
Some energy will be lost due to the work being done against friction upon entry,
but it will be minimal due to the sharp point.

I believe this should be treated as a completely inelastic collision. Mechanical energy will be lost both to friction and to vibrations in the wheel, but that's unimportant here.

 

[haruspex]

a "currency" that is conserved during the exchange is momentum.

Umm, no. It's mounted on a fixed axle, and there will be an unknown reaction from that.

is it correct to take the angular momentum of the dart simply to be about the axis of the wheel,

It's not that the angular momentum is about any particular axis; you choose your axis and calculate the angular momentum about it.

and = r⃗ ×p⃗ \vec r\times \vec p at the instant it hits the rim? (I haven't included the relevant question here.)

Yes, and it is relevant to this question.