Translational and Rotaional motion

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Quincy
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Homework Statement


A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. h' is how far up the smooth side the marble will go, measured vertically from the bottom. Find h', express h' in terms of h.

Homework Equations





The Attempt at a Solution



mgh = mgh' + 1/2 Iw^2 (w = angular velocity)

mgh = mgh' + 1/2 *(2/5)(mR^2)(w^2)

gh = gh' + v^2/5 -- v is the rotational speed of the marble

h' = h - v^2/g -- I can't seem express h' just in terms of h, what am i doing wrong?
 

Answers and Replies

  • #2
kuruman
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The marble will retain the angular speed ω that it has at the bottom of the bowl all the way up to maximum height. In other words, the marble will retain all the rotational kinetic energy that is has at the bottom of the bowl and convert only its translational kinetic energy to gravitational potential energy.

So how will you find the translational kinetic energy at the bottom of the bowl?
 
  • #3
Doc Al
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mgh = mgh' + 1/2 Iw^2 (w = angular velocity)

mgh = mgh' + 1/2 *(2/5)(mR^2)(w^2)
Figure out w by analyzing the first half of the motion separately.
 
  • #4
Quincy
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Figure out w by analyzing the first half of the motion separately.

Thanks, got it now.
 
  • #5
Quincy
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Another problem: A small 15.0 g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 55.0 g and is 90 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 cm/s relative to the table. What is the angular speed (w) of the bar just after the frisky insect leaps?

I thought I should try to figure out the force that the bug applied to the bar, so then i could figure out the torque, but the problem only gives mass and velocity of the bird, which is not enough information to find the force. So I thought I would figure out its kinetic energy.

KE of the bug = 1/2 (0.015 kg)(.2 m/s)^2 = 3 x 10^-4 J

The bar's kinetic energy would equal the kinetic energy of the bird, right?

3 x 10^-4 J = 1/2 (.055 kg)(v)^2

v = .1044 m/s = r*w = (.9)(w) ==> w = .116 rad/s, which is not the right answer... what am I doing wrong?
 
  • #6
Doc Al
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KE of the bug = 1/2 (0.015 kg)(.2 m/s)^2 = 3 x 10^-4 J

The bar's kinetic energy would equal the kinetic energy of the bird, right?
Nope. Why should it? (You mean bug, not bird, I presume.)

Hint: What's conserved? Hint2: The fact that the bar can freely rotate about the nail is important.
 
  • #7
Quincy
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Hint: What's conserved? Hint2: The fact that the bar can freely rotate about the nail is important.

Momentum? But when i use conservation of momentum, i still don't get the right answer.
 
  • #8
Doc Al
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Momentum? But when i use conservation of momentum, i still don't get the right answer.
What kind of momentum? Linear momentum is not conserved.
 
  • #9
Quincy
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What kind of momentum? Linear momentum is not conserved.

Oh, Angular momentum. But still not getting the right answer:

I1*w1 = I2*w2

(.015)(.9^2)(0.2/0.9) = (0.055)(.9^2)(w2)

w2 = 0.061 rad/s....what am i doing wrong now?
 
  • #10
Doc Al
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You have the wrong rotational inertia for the rod.
 
  • #11
Quincy
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Thanks
 
  • #12
heels13
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What is the correct rotational inertia for the bar? Would it be the mass multiplied by half of the bar length squared?
 
  • #13
panchoman
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What is the correct rotational inertia for the bar? Would it be the mass multiplied by half of the bar length squared?

1/3 m L^2
 

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