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Rotational dynamics

"A thin uniform bar of length l and mass M is suspended horizontally at rest. It is suddenly released and at some instant, is struck by a sharp blow vertically upwards at one end (the duration of the impulse is negligible). The impulse causes the bar to rotate at an angular velocity w about the centre of mass."

It then asks what the max height reached by the centre of mass is. I was thinking I should try to find the momentum transferred to the CM and use this to find its initial velocity and then just do projectile motion calculations with the CM? The angular momentum transferred is Iw, but how does this relate to the linear momentum of the bar?
Original post by notastampcollector
"A thin uniform bar of length l and mass M is suspended horizontally at rest. It is suddenly released and at some instant, is struck by a sharp blow vertically upwards at one end (the duration of the impulse is negligible). The impulse causes the bar to rotate at an angular velocity w about the centre of mass."

It then asks what the max height reached by the centre of mass is. I was thinking I should try to find the momentum transferred to the CM and use this to find its initial velocity and then just do projectile motion calculations with the CM? The angular momentum transferred is Iw, but how does this relate to the linear momentum of the bar?


Just to get it clear here.
The metal bar is actually at rest when it is struck? It then spins and moves upwards?
If the bar, after release, has already started falling, there is no way to calculate the max height without knowing how far it has fallen.

To do the problem
The linear and angular momentum are linked by the force F applied by the blow and the short time delta t over which it acts.
For the angular momentum
F(l2)δt=IωF (\frac{l}{2})\delta t = I \omega
you can look up I for a uniform rod.

and for the linear momentum
Fδt=mvF \delta t = mv

This assumes the rod to be stationary initially.

These two equations will allow you to eliminate F and delta t and find v in terms of m, l and w
This is v for the c of m and enables you to find the max height it reaches as you can take its motion to be separate from the rotation.
(edited 13 years ago)
The statement in quotation marks is the question as we received it and I share your confusion about the initial velocity of the rod! I think I shall just assume it is struck immediately after being released (ie it is still stationary at the time of impact).

Thanks for the help. The question then asks how long it takes to pass back through its original position and so I doubled how long it took to reach the maximum height (which gives lw/6g). However the answer supplied is t^2 = 2pi*nl/3g. Where do the pi and n come from?

Thanks again.
Original post by notastampcollector
The statement in quotation marks is the question as we received it and I share your confusion about the initial velocity of the rod! I think I shall just assume it is struck immediately after being released (ie it is still stationary at the time of impact).

Thanks for the help. The question then asks how long it takes to pass back through its original position and so I doubled how long it took to reach the maximum height (which gives lw/6g). However the answer supplied is t^2 = 2pi*nl/3g. Where do the pi and n come from?

Thanks again.


Maybe it means the time taken for the rod to return to its original (horizontal) position, not the c of g. This would explain where the pi and n come from as it depends on the angular velocity. I haven't tried the maths though. It's just the time for n revolutions.
Original post by Stonebridge
Maybe it means the time taken for the rod to return to its original (horizontal) position, not the c of g. This would explain where the pi and n come from as it depends on the angular velocity. I haven't tried the maths though. It's just the time for n revolutions.


I've done it now! Thanks very much for the help! I've been trying to rep you but it just keeps saying I need to share out the repping love before I rep you again. I'll
make a note of it, though, and will rep you when it next lets me!
Reply 5
Avatar for jaroc
jaroc
6
Hm, I keep having trouble working out the expected answer. How did you obtain it?

As the rod rotates about the centre of mass, its angular velocity is constant. We know that dθ=ω dt\mathrm{d} \theta =\omega \, \mathrm{d} t so that time taken for nn revolutions is just τr=2πn/ω\tau_r =2\pi n/\omega and ω=2πn/τr  (∗)\omega =2\pi n/\tau_r\; (*).

The time it takes the rod to move up and back to the original height is, as OP said, τh=ωl/3g  (∗∗)\tau_h =\omega l/3g\; (**).

Let's put (∗)(*) into (∗∗)(**). We obtain:

τh=ωl3g=2πnτrl3g\tau_h =\dfrac{\omega l}{3g}=\dfrac{\dfrac{2\pi n}{\tau_r}l}{3g}

which gives

τhτr=2πn3lg\tau_h \tau_r=\dfrac{2\pi n}{3}\dfrac{l}{g}.

This gives the expected answer only if τh=τr\tau_h=\tau_r, i.e. if the rod has made nn full revolutions up to the moment when its centre of mass is passing back through the original height (and what's more, nn is unspecified!). It doesn't seem to be true for every angular speed ω\omega. If I'm wrong on this point, how could it be proved?
I can only guess that the question means
If the rod passes back through its original position (that is, the exact same place where it started) in a time t, then what would that time be? Then the rod must make n revolutions in the same time it takes the c of g to return to its original position. So the two times would be equal in this case and jaroc's derivation is correct.
Reply 7
Avatar for jaroc
jaroc
6
All right, thanks. I don't like the question then. If τr=τh\tau_r=\tau_h, the time can be expressed in a much simpler way, just by considering the movement of the centre of mass (then τ=ωl/3g\tau=\omega l/3g). Introducing a new unknown nn does not seem to make sense to me as it actually allows to determine nothing! Of course nn can be determined, but then the expression for the time we're trying to work out would simplify back to the form τ=ωl/3g\tau=\omega l/3g. What's the point of introducing nn?...
Original post by jaroc
All right, thanks. I don't like the question then. If τr=τh\tau_r=\tau_h, the time can be expressed in a much simpler way, just by considering the movement of the centre of mass (then τ=ωl/3g\tau=\omega l/3g). Introducing a new unknown nn does not seem to make sense to me as it actually allows to determine nothing! Of course nn can be determined, but then the expression for the time we're trying to work out would simplify back to the form τ=ωl/3g\tau=\omega l/3g. What's the point of introducing nn?...


The point is that the first solution to the problem
τ=ωl/3g\tau=\omega l/3g
is not a unique solution because it depends on omega.
It is not a solution which depends only on g and l.
t can have any value because the original impulse, FδtF \delta t can have any value.
By requiring that the rod also returns to its original horizontal position, we have applied a constraint to the system and made t depend on omega.
We have actually stated that t must be an integral number of revolutions where period t=2πnωt = \frac{2 \pi n}{\omega}
or
ω=2πnt \omega = \frac{2 \pi n}{t}

The solution to the second part of the question, in other words, eliminates omega and produces a series of discrete solutions where n=1,2,3, etc.
This would in practice correspond to discrete values of the applied impulse. The harder the rod was hit, the more times it would spin (n) before returning to its original position. There is no unique value of t for this solution, just a series of larger values for each larger value of n.

This all assumes, of course, that this is what the question actually asked.
(edited 13 years ago)
Reply 9
Avatar for jaroc
jaroc
6
Original post by Stonebridge
The point is that the first solution to the problem
τ=ωl/3g\tau=\omega l/3g
is not a unique solution because it depends on omega.
It is not a solution which depends only on g and l.
t can have any value because the original impulse, FδtF \delta t can have any value.
By requiring that the rod also returns to its original horizontal position, we have applied a constraint to the system and made t depend on omega.
We have actually stated that t must be an integral number of revolutions where period t=2πnωt = \frac{2 \pi n}{\omega}
or
ω=2πnt \omega = \frac{2 \pi n}{t}

The solution to the second part of the question, in other words, eliminates omega and produces a series of discrete solutions where n=1,2,3, etc.
This would in practice correspond to discrete values of the applied impulse. The harder the rod was hit, the more times it would spin (n) before returning to its original position. There is no unique value of t for this solution, just a series of larger values for each larger value of n.

This all assumes, of course, that this is what the question actually asked.


If you put it this way, it does make sense. However, I assumed angular velocity ω\omega to be known and specified in the question. The problem I had with it was that it didn't say anywhere that this velocity corresponded to a number of full revolutions before the c of m of the rod came back to the original height, so that it was impossible to say that the rod could ever be exactly in the original position (both height and angular position as original). For me, introducing nn would be more useful if we wanted to determine the value of impulse needed to allow the rod to return exactly to the original position, but to answer the question it doesn't help much as velocity ω\omega is specified and we cannot do anything about it, only guess that they meant it to correspond to a number of full revolutions. That's why I was surprised by the proposed form of the answer.

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