Steven W. answered • 01/03/17
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Hi Noname!
This problem seems to rely mainly on conservation of energy and conservation of angular momentum, only not in the ways most commonly seen (I think) for those topics. The basic overview here is that lifting the rod gives it gravitational potential energy, which becomes rotational kinetic energy as the rod swings. The swinging rod also picks up angular momentum, which is conserved through the collision with the ball (I will assume that ball is directly below the pivot point of the rod; if it is not, I would need to know where it is with respect to the swing rod). The conservation of angular momentum will allow for the calculation of the "after" quantities (the velocity of the ball after the collision, and the angle to which the rod rises, as its "after" kinetic energy is changed back to gravitational potential energy).
The steps I see for this problem are:
1. Calculate the initial gravitational potential energy of the rod. Once the rod has swung back its straight vertical orientation, this potential energy will have become rotational kinetic energy of the rod (just before it hits the ball).
2. Now that the rotational kinetic energy is known, solve for the angular velocity (of the rod, using the definition that KErot = ½Iω2
3. With ω now known, solve for the angular momentum of the rod just before it strikes the ball. Because the ball is stationary, this is the entire angular momentum pre-collision.
4. The total angular momentum after the collision must equal the angular momentum before (in the absence of net external torque on the system). Both the rod and the ball will have angular momentum (measured around the pivot point of the rod) after the collision, and we can use this to solve for the velocity of the ball.
5. Use the given "after" angular velocity of the rod to compute its "after" rotational kinetic energy.
6. That "after" rotational kinetic energy gets converted back into gravitational potential energy (the reverse of the initial swing-down). This can be used to calculate the angle to which the rod swings.
7. What changes in the elastic case is that the total rotational kinetic energy before gets transferred completely into rotational + translational kinetic energy afterward, and we can use this to solve for the v of the ball and ω of the rod afterwards.
I will post this now, and edit in some details about the procedure shortly.
Steven W.
tutor
Step 2
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The gravitational potential energy from Step 1 gets fully transferred to rotational kinetic energy as the rod reaches the bottom of its swing as is about to hit the ball. So, for the rod of mass m, we have:
KErot = mg(L/2)(1 - cos(60º)) = ½Iω2
where
ω = angular velocity of the rod at the bottom of its swing
I = moment of inertia of a thin rod swung around one end
I = (1/3)mL2 in this case. This is usually given in tables of moments of inertia, but we can derive it if necessary.
So we have:
mg(L/2)(1 - cos(60o)) = (1/2)[(1/3)mL2](ω2)
Solve this expression for ω.
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01/03/17
Steven W.
tutor
Step 3
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Once ω is known, compute the angular momentum of the rod at the bottom of its swing.
L = Iω
Both I and ω are known from the previous step. The angular momentum of the rod constitutes the entire initial angular momentum of the collision, since the ball starts at rest.
Step 4
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Write an expression for the total angular momentum after the collision. At this point, both the ball and the rod have angular momentum (measured around the rotation point of the rod).
For the rod, after the collision:
Lra = Iωa (with I the same as before, and ωa given in the problem)
For the ball after the collision:
This can be a little trickier to see, since we usually do not think of objects moving in a straight line as having angular momentum. But it turns out they do, when measured around a point off their line of travel. It turns out to be:
L = mvr
where
m = mass of object
v = speed of object
r = perpendicular distance from the point of rotation (off-line) to the line of travel of the object
For the case of the ball, we want to measure its angular momentum around the same axis of rotation as for the rod (so that we can just add the two angular momenta as numbers). Taking the rotation point of the rod as the axis, the perpendicular distance from that point to the line of motion of the ball (presuming the ball is directly below the rotation point, at the same height as the rod when it hangs vertically), is just L, the length of the rod.
Then:
Lb = mbvL
where
mb = mass of the ball
v = (linear) speed of the ball after the collision
Then, the total angular momentum after the collision is:
Lafter = Lra + Lb = angular momentum before the collision (by conservation of angular momentum), which was calculated in Step 3
So:
Iωa + mbvL = Iω
With everything given above and in previous steps, the only unknown is v, the velocity of the ball. You can thus solve for v in this step, answering the first question.
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01/03/17
Steven W.
tutor
Step 5
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Calculate the rotational kinetic energy of the rod after the collision:
KErot-a = (1/2)Iωa2
(all of these quantities are known at this point)
This rotational kinetic energy gets converted back into gravitational potential energy as the rod's center of gravity rises.
Step 6
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This involves the same expression for the rise of the rod's center of gravity as in Step 1. In this case, though, the initial angular velocity of the rod (and thus its initial rotational kinetic energy, calculated in Step 5) is known, and the angle to which the rod rises is unknown. But you can use the same relationship to solve for the angle (θa):
(1/2)Iωa2 = mg(L/2)(1 - cos(θa))
Everything in this expression is known except θa. So you can solve for θa here, answering the second question (note that taking inverse cosine will be involved in the solution).
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01/03/17
Steven W.
tutor
Step 7
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For the last question, we no longer know ωa, so the conservation of angular momentum equation now has two unknowns: v and ωa. So we cannot solve for either one with just that equation. We need a second one, which is provided by the statement that the collision is elastic.
Because of that, we know the total kinetic energy (rotational + translational) after the collision equals the total rotational kinetic energy before the collision, which we calculated back in Step 2 (that result does not change because of this). So:
(1/2)Iω2 = KErot-a (for the rod) + KEtrans-a (for the ball) = (1/2)Iωa2 + (1/2)mbv2
When combined with the conservation of angular momentum equation, this constitutes two equations with two unknowns (ωa and v), so you can then solve for both using those equations.
v is one of the quantities you need to answer the last question. Then you just have to convert ωa into the linear speed at the end of the rod. Since the end of the rod is traveling in a circle of radius L, we can relate the tip's translational velocity to its rotational velocity using:
va = Lωa
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01/03/17
Steven W.
tutor
I hope this helps get you on your way! There is a lot going on here, so, if you have any questions about any of the steps at all, or if you would like to check results, just let me know.
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01/03/17
Steven W.
01/03/17