Which Rolling Object Reaches the Bottom of the Ramp First?

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A solid ball, hollow ball, and hoop of the same radius are released simultaneously down a ramp, and the order in which they reach the bottom is determined by their moments of inertia (MOI). The solid ball has the lowest MOI (2/5ma^2), followed by the hollow sphere (2/3ma^2), and the hoop has the highest MOI (ma^2). As a result, the solid ball reaches the bottom first, followed by the hollow sphere, and then the hoop. The discussion also touches on the relationship between torque, angular momentum, and Newton's Second Law, emphasizing that torque is the rate of change of angular momentum. The confinement of motion to a plane is explained through the conservation of angular momentum, which maintains the initial plane of motion.
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Homework Statement



A solid ball of constant density, a hollow ball and a hoop have the same radius.
They are simultaneously released to roll without slipping down the same ramp. Write
the order in which they reach the bottom and justify your answer.


Homework Equations





The Attempt at a Solution



The thing is, it's worth 7 marks..I have no idea what working i need to do!

So as they fall, PE is converted into KE..

KE is translational and rotational..

so mgh = 1/2mv^2 + 1/2Iw^2 = 1/2mv^2 + 1/2 I (v/r)^2 where h is the height fallen

so gh = 1/2v^2 + 1/2 I/m (v/r)^2...

Now I am guessing the I/m will be the most for the solid ball, then for the hollow ball and least for the hoop..so hoop arrives first, then hollow ball, then solid ball..

is this right? How can i justify it more rigorously?
 
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bon said:
Now I am guessing the I/m will be the most for the solid ball, then for the hollow ball and least for the hoop..
Why are you guessing? Look it up! (Or just think about it a bit.)

Other than that, you have the right idea.
 


Doc Al said:
Why are you guessing? Look it up! (Or just think about it a bit.)

Other than that, you have the right idea.

Further to this (sorry it was ages ago...)

I think the MOI of a solid sphere would be 2/5ma^2, of a hollow sphere ma^2, and of a hoop ma^2..

So solid ball arrives first then hoop and hollow sphere arrive together
 


Is this right?
 


bon said:
Further to this (sorry it was ages ago...)

I think the MOI of a solid sphere would be 2/5ma^2, of a hollow sphere ma^2, and of a hoop ma^2..

So solid ball arrives first then hoop and hollow sphere arrive together
You're almost there. You have the correct MOI for the solid sphere and for the hoop, but not for the hollow sphere.
 


Doc Al said:
You're almost there. You have the correct MOI for the solid sphere and for the hoop, but not for the hollow sphere.


But isn't all the mass situated on the outer shell so for each infinitesimal mi the distance to the central axis is ri^2? so MOI is ma^2?
 


bon said:
But isn't all the mass situated on the outer shell so for each infinitesimal mi the distance to the central axis is ri^2? so MOI is ma^2?
Yes, the mass is on the outer shell. But the distance of that shell to the axis varies from 0 to R. (The distance to the center is fixed, but not to the axis.)
 


Doc Al said:
Yes, the mass is on the outer shell. But the distance of that shell to the axis varies from 0 to R. (The distance to the center is fixed, but not to the axis.)

Oh i see so 2/3ma^2. Great thanks.

so order: solid sphere first, then hollow, then hoop...?
 


bon said:
Oh i see so 2/3ma^2. Great thanks.

so order: solid sphere first, then hollow, then hoop...?
Right! The more energy (per unit mass) required to turn the thing, the less energy is left for translational motion.
 
  • #10


Doc Al said:
Right! The more energy (per unit mass) required to turn the thing, the less energy is left for translational motion.


Yay...thanks for all your help!

Sorry to corrupt this thread with a different question (but i guess it's sort of related)..

A rigid body is free to rotate about a fixed axis k through the origin. A force F
is applied to the body at a point R. State the relationship between the torque due to
F and the angular momentum about the axis. How is this related to Newton’s Second
Law applied to linear momentum?

So i know torque due to F is the rate of change of angular momentum i.e. G = dJ/dt

or here: R x F = d/dt(mR x R dot)

But what about the second part? how is this related to NII for linear momentum..

I know that if you start with J = m r x r dot and differentiate you see that dJ/dt = r x mr double dot..then you can replace mr double for F due to NII, but i don't think this answers the question...Thanks

Also - when you have a point mass m at a radius r in a gravitational potential V(r), why is the general motion confined to a plane..I know that the angular momentum about z i.e. r x r dot is constant..but why does this => plane? thanks again!
 
  • #11


bon said:
Yay...thanks for all your help!

Sorry to corrupt this thread with a different question (but i guess it's sort of related)..

A rigid body is free to rotate about a fixed axis k through the origin. A force F
is applied to the body at a point R. State the relationship between the torque due to
F and the angular momentum about the axis. How is this related to Newton’s Second
Law applied to linear momentum?

So i know torque due to F is the rate of change of angular momentum i.e. G = dJ/dt

or here: R x F = d/dt(mR x R dot)

But what about the second part? how is this related to NII for linear momentum..

I know that if you start with J = m r x r dot and differentiate you see that dJ/dt = r x mr double dot..then you can replace mr double for F due to NII, but i don't think this answers the question...Thanks
I really don't know what they are looking for with that second question. I would have done something similar to what you did. Start with the definition of angular momentum, differentiate and apply NII. Beats me!

Also - when you have a point mass m at a radius r in a gravitational potential V(r), why is the general motion confined to a plane..I know that the angular momentum about z i.e. r x r dot is constant..but why does this => plane?
The initial plane of motion can be defined as that perpendicular to the vector ri X vi. But angular momentum conservation implies that that r X v doesn't change, so the plane is fixed.

Look at it this way. The vectors r, v, and the acceleration start out in the same plane, right? So how can the mass leave that plane?
 
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