Noname N.

asked • 01/02/17

Rotational physics problem

a 1.5kg uniform rod of length .75 meters swings from rest around an axis through its end through an angle of 60 degrees. It strikes the center of a small .25 kg ball which sits on the edge of a table.
 
After the collision, the rod has angular velocity of 2.5 rad/s. what is the velocity of the ball after the equation?
 
What maximum angle does the rod swing through after the collision?
 
If instead the rod and the ball collide elastically, find the linear velocity of the end of the rod and the velocity of the ball immediatly after the collision .

1 Expert Answer

By:

Steven W.

tutor
I apologize that some of this may seem pretty opaque without a diagram, but here goes:
 
Step 1
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We can measure the gravitational potential energy of the rod (an extended body) by tracking the height of its center of mass.  Because the rod is uniform, its center of mass is at its midpoint.  Let's measure the potential energy relative to when the rod is hanging vertically.  At this point, the center of gravity is at (L/2) below the pivot point, where L is the length of the rod.  Let's take this height of the center of gravity as "h = 0."
 
The question then is, how high is the center of mass above that level when the rod is held at 60º.  I am presuming this is 60º with respect to the vertical.  If so, then I would argue that the vertical height of the center of gravity above its zero position at that angle is (L/2)(1 - cos(60º)).  See if you can demonstrate to yourself -- perhaps with the help of a diagram -- that this is true.  The gravitational potential energy of the rod at that point is then mg(L/2)(1 - cos(60o)).
 
 
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01/03/17

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 = Iω (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:
 
a + 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

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