• ledphones
In summary, a system consisting of two 2.1 kg balls attached to a thin rod of negligible mass and 63 cm in length is initially horizontal. A 50 gram wad of wet putty with a speed of 3.9 m/sec drops onto one of the balls and sticks to it. After the collision, the angular speed of the system is 0.146 rad/sec and the ratio of the kinetic energy of the entire system just after the collision to just before the collision is 0.01175. To determine the angle the system will rotate until it momentarily stops, conservation of energy is used. The potential energy of the system is set equal to the kinetic energy, and the height is solved for. The calculated height
ledphones

Homework Statement

Two 2.1 kg balls are attached to the ends of a thin rod of negligible mass, 63 cm in length. The rod is free to rotate in a vertical plane about a horizontal axis through its center. With the rod initially horizontal as shown, a 50 gm wad of wet putty drops onto one of the balls with a speed of 3.9 m/sec and sticks to it. a) What is the angular speed of the system just after the putty wad hits?

.146 OK

b) What is the ratio of the kinetic energy of the entire system just after the collision to just before the collision?

KEafter/KEbefore = *
.01175 OK

c) Through what angle will the system rotate until it momentarily stops?

The Attempt at a Solution

Im not sure how to approach part c since I don't know the acceleration or the time it takes to stop.

Have you thought about using a conservation of energy approach?

how would you use that to solve for the angle? would you have
K+U=K+U
1/2 I w^2 + 0 = 0 + ? what would the potential energy be here? and how could you relate it to the angle (or distance traveled?) would it be through a kinematic equation?

ledphones said:
how would you use that to solve for the angle? would you have
K+U=K+U
1/2 I w^2 + 0 = 0 + ? what would the potential energy be here? and how could you relate it to the angle (or distance traveled?) would it be through a kinematic equation?

Consider that the rod and balls without the wad of putty are symmetrical, so that they would be in equilibrium at any angle. No change in PE occurs if its angle is changed. If it were set rotating, it would do so at a constant velocity (barring friction, of course).

The wad sticking to one end changes things. There will be a PE associated with its relative height. It's the only part of the system for which a net exchange of energy with the gravitational field will occur.

so you'd know the potential energy once the wad hits is (m+M)gh. so i'd set that equal to 1/2 Iw^2 and solve for h. but this was a value of .0002133 which is much to small.

ledphones said:
so you'd know the potential energy once the wad hits is (m+M)gh. so i'd set that equal to 1/2 Iw^2 and solve for h. but this was a value of .0002133 which is much to small.

Why m+M?

i tried just M now. and that was wrong as well. as for my reason for adding those two i though you had to add the mass of the wad of gum plus one of the balls

ledphones said:
i tried just M now. and that was wrong as well. as for my reason for adding those two i though you had to add the mass of the wad of gum plus one of the balls

I see. Well, as I said, the rod and balls are symmetrical, so their contribution is nil except that they add to the moment of inertia of the whole.

You'll have to show your calculation, reasoning, and numbers if we're to see what's going wrong.

ok so i tried:
1/2(MR^2+MR^2+mr^2)w^2=mgh
h=(.5(2*2.1*.351^2+.050*.351^2)*.146^2)/(.05*9.8)

ledphones said:
ok so i tried:
1/2(MR^2+MR^2+mr^2)w^2=mgh
h=(.5(2*2.1*.351^2+.050*.351^2)*.146^2)/(.05*9.8)

Okay. The rod has length 63cm, so the radius is 0.315m. I think you've got a typo there where you've used 0.351.

The height that's calculated will be the distance above the initial height (rod horizontal) that the wad must reach. Immediately after impact the wad is heading downward and picking up speed, maxing out at the bottom of its arc. It then proceeds up the other side, losing speed, until the rod is once again horizontal and the speed is the same as the initial speed, and the KE is the same as the initial KE. After this it continues upwards until it reaches height h above the horizontal.

Ok so now I tried

1/2Mv^2/r^2=Mgh

h=.5(2.1)(3.9^2)/(.315^2*9.81*2.1)

this was wrong :(

Last edited:
ledphones said:
Ok so now I tried

1/2Mv^2/r^2=Mgh

h=.5(2.1)(3.9^2)/(.315^2*9.81*2.1)

I don't know what to make of that; the units on the LHS of the first expression don't come out as energy. What's the equation meant to describe?

Let's go back to what is known. What value did you calculate for the KE immediately after the collision?

.0044945452(from the following expressions:)

1/2Iw^2
I=(2*MR^2+mR^2)

Last edited:
ledphones said:
.0044945452(from the following expressions:)

1/2Iw^2
I=(2*MR^2+mR^2)

Okay. That's about what I calculate: 4.474 x 10-3 J. I suppose I'm keeping a few more decimal places in all the intermediate results.

Now, the ball with the wad travels around in its arc attached to the rod end until the rod is once again horizontal. That's an angle of pi radians so far traveled. At this point the system has the same KE as it had when the wad of putty first struck. As the wad rises above the horizontal, the system loses KE to gravitational PE.

So how high must the wad of putty rise in order to "steal" all of the rotational KE?

h = KE/(m*g)

What value do you get for h and the resulting angle above horizontal? (an angle which must be added to the angular distance already traveled by the ball and wad).

Attachments

• Fig1.gif
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1. How does the mass of the putty wad affect its rotational collision?

The mass of the putty wad affects its rotational collision by determining the amount of rotational inertia it has. A heavier putty wad will have more rotational inertia and therefore require more force to change its rotational motion.

2. What role does the shape of the putty wad play in its rotational collision?

The shape of the putty wad affects its rotational collision by determining its center of mass and its moment of inertia. A putty wad with a more spread out shape will have a larger moment of inertia and be more difficult to rotate compared to a more compact putty wad with the same mass.

3. How does the surface on which the putty wad collides affect its rotational motion?

The surface on which the putty wad collides can affect its rotational motion by providing friction or a lack thereof. A rough surface will provide more friction, causing the putty wad to slow down and possibly change its rotational direction. A smooth surface will provide less friction, allowing the putty wad to maintain its rotational motion.

4. Can the speed of the putty wad affect its rotational collision?

Yes, the speed of the putty wad can affect its rotational collision. A higher speed will result in a greater momentum and therefore a stronger impact on the object it collides with, potentially causing a larger change in its rotational motion.

5. Is there a relationship between the angle of impact and the resulting rotational motion of the putty wad?

Yes, the angle of impact can affect the resulting rotational motion of the putty wad. A direct impact at a 90-degree angle will result in the greatest change in rotational motion, while a glancing impact at a smaller angle will result in a smaller change in rotational motion.

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