Angular Momentum: Rod on Rotating Disk

In summary: How much energy is lost due to friction?In summary, a uniform disk and a non-rotating rod of the same mass and diameter rotate around a frictionless spindle with a frequency of 5.2 rev/s. In the idealized case, where the rod sticks tightly to the disk upon impact, the combination turns with a frequency of 3.12 rev/s. In the more realistic case, where there is a frictional force between the two surfaces, the constant C is equal to 0.667. The disk slows down and the rod speeds up until they are moving together, with a frequency f'. To find f', we must consider the net external torque on the system, which is equal to zero. This implies that
  • #1
nastassja
6
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Homework Statement



A uniform disk (like a record turntable) turns with a frequency of 5.2 rev/s around a frictionless spindle. A non-rotating rod of the same mass as the disk and length equal to the disk's diameter is placed on the freely spinning disk so both turn around the spindle. (Idisk = M R^2/2, Irod = M L^2/12)

a) First consider the idealized case where the rod sticks tightly to the disk immediately upon impact. How fast does the combination turn?

f = rev/s 3.12 OK

b) More realistically, there is a frictional force between the two surfaces such that the magnitude of the torque slowing down the disk is equal to the magnitude of the torque speeding up the rod (this is a consequence of Newton's 3rd Law.). Hence we have adisk = -C arod, where a generically means angular acceleration. Find the constant C.

C = .667 OK




c) The disk slows and the rod speeds up until they are moving together. Find the frequency with which the combination turns in this case.

f' = ?

HELP: Consider the net external torque on the system. What is it? What does this imply?

d) Suppose Idisk = .65 kg m2. What is the magnitude of the kinetic energy lost due to friction?

|Wfric| = ?

HELP: Use the work-energy theorem: the work done by friction is equal to the change in kinetic energy.

Homework Equations



L=Iw
w=2pi*f
torque=Ia



The Attempt at a Solution



I got a and b and I'm now stuck on c. I know I can figure out d if I have c. I don't understand how to work this one at all. All I need is some help starting it.
 
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  • #2
Hi I have a very similar problem. How did u get the first to parts solving for F and C?
 
  • #3
c) What is the condition for the conservation of angular momentum? Does it hold in this case?

d) Compare the rotational KE of the system before and after the rod is placed on the turntable.
 

1. What is the concept of "Angular Momentum" in relation to a rod on a rotating disk?

Angular momentum is a physical quantity that measures the rotational motion of an object. In the case of a rod on a rotating disk, it refers to the momentum of the rod as it spins around the center of the disk.

2. How is angular momentum calculated for a rod on a rotating disk?

The angular momentum of a rod on a rotating disk can be calculated using the formula L = Iω, where L is the angular momentum, I is the moment of inertia of the rod, and ω is the angular velocity of the disk.

3. What is the relationship between angular momentum and rotational inertia?

Angular momentum and rotational inertia are directly proportional to each other. This means that an increase in rotational inertia will result in a corresponding increase in angular momentum, and vice versa.

4. How does the distribution of mass affect the angular momentum of a rod on a rotating disk?

The distribution of mass along the length of the rod can greatly affect the angular momentum. A more concentrated mass towards the center of the rod will result in a higher angular momentum, while a more evenly distributed mass will result in a lower angular momentum.

5. Can the angular momentum of a rod on a rotating disk be changed?

Yes, the angular momentum of a rod on a rotating disk can be changed by altering its rotational speed or by changing the distribution of mass along the rod. Additionally, external forces such as friction can also affect the angular momentum of the system.

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