Angular acceleration, density question

AI Thread Summary
The discussion centers on two physics questions involving angular acceleration and density. For the first question, the calculation of angular acceleration for a DVD modeled as a thin uniform cylinder yields an incorrect result due to the moment of inertia being misapplied; the correct formula for a thin cylinder is I = 1/2 Mr². The second question involves determining the density ratio of ice to water using Archimedes' principle, but participants note that the questions should ideally be separated for clarity. There is also confusion regarding the terminology used to describe the DVD's shape, with suggestions that "thin cylinder" may not be the best descriptor. Overall, the thread highlights the importance of precise definitions and correct application of physics principles in problem-solving.
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Homework Statement



First question:
"A DVD player can exert a torque on a DVD of 10−3 N⋅m. What is the DVD’s angular acceleration in rad/s2?

Details and assumptions
  • The DVD has a mass of 0.015 kg and a radius of 0.06 m.
  • A DVD can be modeled as a thin uniform cylinder.
"

Second question:
"A cube of ice 2 cm on a side is floating in a glass of water. You observe that the top of the ice is 0.166 cm above the surface of the water. What is the ratio of the density of ice to the density of water?"

Homework Equations



First question:
α = \tau / I
I = Ʃmr2

The Attempt at a Solution



First question:
α = 0.001 / ((0.015)(0.06)2)
= 18.5185

Not sure how I can do the second question at all...
I've got these two questions from the internet.
Please help me solve these question.
 
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letsfailsafe said:
α = 0.001 / ((0.015)(0.06)2)
= 18.5185
Yes, given the problem statement, but I'm astonished a DVD is to be modeled as a thin cylinder. A disc would be more reasonable, and that would change the answer.

For the 2nd problem (in future, pls use separate threads for unrelated problems), what determines the volume of the ice that will be below water?
 
α = 0.001 / ((0.015)(0.06)2)
= 18.5185

is not right...
 
1. The moment of inertia of a disc or thin cylinder is

I = 1/2 Mr2

so you are out by a factor of 2.

2. For this question you can start out with Archimedes' principle which states that the bouyant force on an object is equal to the weight of the liquid that it displaces.
 
Basic_Physics said:
1. The moment of inertia of a disc or thin cylinder is
I = 1/2 Mr2
For a disc, yes, but not for a thin cylinder about its axis. And it does say cylinder, strange though that is.
Perhaps it meant short fat cylinder, but then why not just say disc?
 
A thin solid cylinder would be a disc?
 
Maybe I usually help 'too much'... :redface: To get started, think about the forces acting on the ice cube.
ƩF=0 ⇔
G=N...
 
Basic_Physics said:
A thin solid cylinder would be a disc?
It doesn't say solid, you've added that. The most obvious interpretation of 'thin cylinder' is that it is a tube with a thin wall, i.e. the cross-section has a narrow annulus. If it means thin the other way, why not say a very short solid cylinder, or even more naturally, a disc? But it is quite possible that the question is just very poorly worded.
 

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