moment of inertia

[toki]

a ring of inner Radios r1 and outer Radios r2 is falling(free fall) from a plane, which is angled (theta) with the horizontal axis.

what will be itz final valocity?

(plz ans it using the moment of inertia for ring:(.5*m(r1^2+r2^2))

[swansont]

What work have you done so far?

[toki]

yeah....

[J.C.MacSwell]

a ring of inner Radios r1 and outer Radios r2 is falling(free fall) from a plane, which is angled (theta) with the horizontal axis.

what will be itz final valocity?

(plz ans it using the moment of inertia for ring:(.5*m(r1^2+r2^2))

Are you sure that is the right context of the question?

 

Free fall does not take place on an angled plane.

[swansont]

yeah....

 

 

This is homework help, not the "do the homework for you" section. How would you normally approach solving an object on an inclined plane problem?

[toki]

I mean that the ring is not thrown in any initial valocity.just as you have put the ring top of the plane and escaped it!

and this is for physics olympiad.not my homework!!

I have got the solution for a sphere or cylinder, in the physics by robert resnic.but cannt solve this problem for a ring.

[swansont]

I mean that the ring is not thrown in any initial valocity.just as you have put the ring top of the plane and escaped it!

and this is for physics olympiad.not my homework!!

I have got the solution for a sphere or cylinder, in the physics by robert resnic.but cannt solve this problem for a ring.

 

Then you should be able to work through this. What principle do you apply to this sort of problem?

[toki]

for the height of the plane 'h'

mgh=0.5*I*w^2 +0.5*m*v^2

=.5*(.5*mr^2)*(v^2/r^2)+.5*mv^2

solving this give

v=root of (4/3 * gh)[for a cylinder ]

but for a ring there are tro Radios

inner and outer.so how I can solve it for a ring?

[swansont]

for the height of the plane 'h'

mgh=0.5*I*w^2 +0.5*m*v^2

=.5*(.5*mr^2)*(v^2/r^2)+.5*mv^2

solving this give

v=root of (4/3 * gh)[for a cylinder ]

but for a ring there are tro Radios

inner and outer.so how I can solve it for a ring?

 

 

Which radius is the one that you want to use to convert rotational speed into linear speed? (I.e. what part of the wheel is touching the surface, so that a revolution can tell you the distance traveled?)

 

For the moment of inertia substitution, both radii will show up.

[toki]

 

 

Which radius is the one that you want to use to convert rotational speed into linear speed? (I.e. what part of the wheel is touching the surface, so that a revolution can tell you the distance traveled?)

 

For the moment of inertia substitution, both radii will show up.

r1 is the outer one.

but if I want to find the moment of inertia I must have to use this two radios r1 and r2,isn't it?

if I do it,in the last solution I will find a ans with g,h,r1and r2.

because r1 and r2 cannot be cut up with the r^2 of w^2=v^2/r^2

[swansont]

r1 is the outer one.

but if I want to find the moment of inertia I must have to use this two radios r1 and r2,isn't it?

if I do it,in the last solution I will find a ans with g,h,r1and r2.

because r1 and r2 cannot be cut up with the r^2 of w^2=v^2/r^2

 

Yes. You gave the moment of inertia in your first post. It has both values in it, and they will both show up in the answer.

 

Is the problem in deriving the moment of inertia?

[imatfaal]

Just out of curiosity - if the only force on the ring is gravity why is it rotating?

[DrP]

Because it is rolling down the slope imat'

[swansont]

Just out of curiosity - if the only force on the ring is gravity why is it rotating?

 

 

The standard assumption is that it rolls without sliding. So there's friction to get it rolling but none that robs the system of energy.

[imatfaal]

Because it is rolling down the slope imat'

 

Which it wouldn't do without friction - in the (admittedly confused) OP the term "freefall" is used which is normally taken to mean only under force of gravity. As JCMac pointed out this was an odd formulation for an inclined slope; obviously you need the normal reaction for the slope to be taken into account. But including friction as well...

 

But in getting ready for a physics competition I would have thought that the use of the term freefall should actually mean something - and the set of friction coefficients to get a perfect roll with no sliding (per SwansonT's message above) should be an explicit part of a question. It would not be beyond the bounds of teachers and quiz setters to deliberately fool competitors by asking a trick question for a sphere, a cylinder, and a ring and thus tempting students into assuming rolling (cos of the different moments of inertia) when they should be assuming sliding