Rolling bodies - mechanics

Original post by newblood

However, the diagram below then says the ''moment of inertia is taken about the point of contact by the parallel axis theorem''. If i have understood, this right then that is a contradiction of above. Also, in a STEP question i just did about rolling bodies, the first part involved finding the M.I. about a point on circumference of hoop so i am inclined to believe that the second definition is correct (unless they somehow both are?)

Could anyone clarify this for me please

Bit rusty on this, but I can't see the bit you're refering to.

All I can see is the "is related to the moment of inertia about the point of contact by the parallel axis theorem" next to the diagram.

Reply 2

OP

Original post by ghostwalker

Bit rusty on this, but I can't see the bit you're refering to.

All I can see is the "is related to the moment of inertia about the point of contact by the parallel axis theorem" next to the diagram.

Yeah thats what i mean. What does related to mean if it isnt implying that is the axis of rotation for the moment of inertia?

Reply 3

Original post by newblood

Yeah thats what i mean. What does related to mean if it isnt implying that is the axis of rotation for the moment of inertia?

Just means that the moment of inertia about the point of contact is related to the moment of inertia about the centre of mass, via the parallel axis theorem.

Reply 4

OP

Original post by ghostwalker

Just means that the moment of inertia about the point of contact is related to the moment of inertia about the centre of mass, via the parallel axis theorem.

so what is the point of mentioning it unless the moment of inertia that should be used to calculate KE is about the point of contact?

sorry if im not being clear: but can you explicitly tell me, in an example where a ball is rolling on the surface, to calculate the KE, about what axis is the moment of inertia specified for (that is used in calciulating the rotational KE). And is this the same when it has nearly falled off a surface, so on the point of contact and slipping is not occurring

thank you

Reply 5

Original post by newblood

so what is the point of mentioning it unless the moment of inertia that should be used to calculate KE is about the point of contact?

sorry if im not being clear: but can you explicitly tell me, in an example where a ball is rolling on the surface, to calculate the KE, about what axis is the moment of inertia specified for (that is used in calciulating the rotational KE). And is this the same when it has nearly falled off a surface, so on the point of contact and slipping is not occurring

thank you

Not sure if I can be clear, as I said, I'm rather rusty on this.

At any moment the point of the ball in contact with the ground is stationary, and we can consider this to be a pivot, and use the MofI about that point in working out the total KE (there is in fact only rotational KE about that point).

Alternatively we can work out the rotational energy about the centre of mass and add the translational KE, to get the total KE. They are one and the same - I think!

Not sure what you mean by nearly falling off a surface.

Reply 6

OP

Original post by ghostwalker

Not sure if I can be clear, as I said, I'm rather rusty on this.

At any moment the point of the ball in contact with the ground is stationary, and we can consider this to be a pivot, and use the MofI about that point in working out the total KE (there is in fact only rotational KE about that point).

Alternatively we can work out the rotational energy about the centre of mass and add the translational KE, to get the total KE. They are one and the same - I think!

Not sure what you mean by nearly falling off a surface.

rolling off the edge of the table is what i meant

thanks for your help anyway, hopefully someone who has studied this level of mechanics recently will pass by

Reply 7

(edited 10 years ago)

Reply 8

Original post by newblood

And is this the same when it has nearly falled off a surface, so on the point of contact and slipping is not occurring

thank you

It's not clear what you mean here. If a ball/cylinder reaches the edge of a table, and its point of contact is still not slipping, that *at that instant*, the energy situation is exactly as before it reached the edge.

However, thereafter, the object is rotating without slipping about the edge of the table under the influence of the torque generated by its weight through its c.o.g. and the reaction at the edge (which won't be vertical), so this situation is far trickier to analyse, as the object would have some varying angular acceleration. In fact, I suspect that this scenario would be very challenging even for a STEP problem.

Reply 9

OP

Original post by atsruser

It's not clear what you mean here. If a ball/cylinder reaches the edge of a table, and its point of contact is still not slipping, that *at that instant*, the energy situation is exactly as before it reached the edge.

However, thereafter, the object is rotating without slipping about the edge of the table under the influence of the torque generated by its weight through its c.o.g. and the reaction at the edge (which won't be vertical), so this situation is far trickier to analyse, as the object would have some varying angular acceleration. In fact, I suspect that this scenario would be very challenging even for a STEP problem.

In the case of a cylinder you have shown that the KE is rotatoional ke about cog and translational KE which is also the same as its rotational KE about the point of contact. If we now had a different object,say a solid sphere would we use the 2nd equation that KE is purely rotational about point of contact and in this case it isnt equal to rotational ke of cog and translational KE because of the form of the MI?

Reply 10

Original post by newblood

In the case of a cylinder you have shown that the KE is rotatoional ke about cog and translational KE which is also the same as its rotational KE about the point of contact. If we now had a different object,say a solid sphere would we use the 2nd equation that KE is purely rotational about point of contact and in this case it isnt equal to rotational ke of cog and translational KE because of the form of the MI?

The whole argument is the same regardless of the specific rolling object. I meant to point that out at the end, but I forgot.

So the result is indeed still true if you have a rolling sphere rather than a cylinder.

Reply 11

OP

Original post by atsruser

The whole argument is the same regardless of the specific rolling object. I meant to point that out at the end, but I forgot.

So the result is indeed still true if you have a rolling sphere rather than a cylinder.

Oh yeas i see now in your algebra that it applies for all shapes as you have merely used I for the object rather than specifying it - i read the text but skimmed through the algebra before as i was in a rush.

Thanks for the very informative reply - has cleared it all up now!

One question though: In this STEP question below. When you consider the K.E. of the hoop at the edge, how is it any different from when it is on the surface. I imagine it is supposed to be as i tried using conservation of energy and and resolving radially (to make use of R=0 when sphere loses contact), but i cant see what is different about the K.E. when it has rolled over the edge.

Reply 12

Original post by newblood

Oh yeas i see now in your algebra that it applies for all shapes as you have merely used I for the object rather than specifying it - i read the text but skimmed through the algebra before as i was in a rush.

Thanks for the very informative reply - has cleared it all up now!

One question though: In this STEP question below. When you consider the K.E. of the hoop at the edge, how is it any different from when it is on the surface. I imagine it is supposed to be as i tried using conservation of energy and and resolving radially (to make use of R=0 when sphere loses contact), but i cant see what is different about the K.E. when it has rolled over the edge.

When the hoop is making an angle

\theta

with the vertical, then its c.o.g has fallen a certain distance from its height when on the surface. This means that it has lost PE, which will have been converted into rotational KE, since the hoop, by supposition, is rotating without slipping about the edge. The rotational velocity of the hoop is increasing with time (or angle, if you prefer) as the c.o.g. falls lower and lower.

The rotational KE of the hoop when rotating about the edge is given by

\frac{1}{2}I\omega^2_\theta

where

\omega_\theta

is angular velocity of the hoop about the edge, when it is at angle

\theta

. Note that the MI

I

here is the same one as we used when it is rolling on the surface, since the hoop is still rotating about a fixed point on its circumference.

Also, if

v_\theta

is the tangential velocity of the c.o.g at angle

\theta

, then

v_\theta = a \omega_\theta

, since the c.o.g. is at the centre of the hoop of radius

a

.

I have to say that I've had a hard time convincing myself of the underlying physics of this question, in the sense of trying to figure out from physical principles why the hoop doesn't leave the edge immediately. In fact, I think for some sufficiently large value of rolling speed

v

, it will leave the edge immediately, due to their being insufficient component of

g

towards the edge to keep it in contact.

Reply 13

Original post by atsruser

I have to say that I've had a hard time convincing myself of the underlying physics of this question, in the sense of trying to figure out from physical principles why the hoop doesn't leave the edge immediately.

You have, in essence, motion in a circle, with the edge being the centre of the circle. As long as the force required to move in a circle about that edge is less than mg, it won't leave the edge immediately.

Reply 14

Original post by ghostwalker

You have, in essence, motion in a circle, with the edge being the centre of the circle.

The question as stated presupposes the circular motion. I was trying to see how it would arise and be maintained, from first principles, given that the hoop starts with a horizontal component of velocity of magnitude

v

. In order to maintain contact with the edge, the hoop has to be accelerated towards the edge. This is essentially a kinematic view of the problem.

(In fact, this acceleration must come from the

mg\cos\theta

component of

mg

. At some point, this becomes too small to give the hoop the necessary acceleration towards the edge).

As long as the force required to move in a circle about that edge is less than mg, it won't leave the edge immediately.

I'm not sure where you get

mg

from.

The equation of motion resolved towards the edge is

\displaystyle mg\cos\theta -R =\frac{mv^2_\theta}{a}

.

When the hoop is just at the edge

\theta = 0

and

mg\cos\theta=mg

which is its maximum value. As

\theta

increases

mg\cos\theta

decreases, while

v_\theta

increases from its initial value of

v

.

For a while,

\displaystyle mg\cos\theta > \frac{mv^2_\theta}{a}

and we need a non-zero

R

to reduce the radial force to exactly that needed to generate rotational speed

v_\theta

.

But since

mg\cos\theta

is decreasing, and

v_\theta

is increasing, at some point, we have

\displaystyle mg\cos\theta = \frac{mv^2_\theta}{a}

i.e. the radial component of

mg

is exactly enough to generate rotational speed

v_\theta

and at this point

R

becomes 0 - it was decreasing all the time that

mg\cos\theta

and

\displaystyle mv^2_\theta/a

were becoming closer.

Thereafter, the hoop has left the edge and we no longer have circular motion.

Reply 15

Original post by atsruser

I'm not sure where you get

mg

from.

From theta= 0. My post refered to leaving the edge immediately, which is what the post that I responded to was refering to.

(edited 10 years ago)

Reply 16

OP

Original post by ghostwalker

You have, in essence, motion in a circle, with the edge being the centre of the circle. As long as the force required to move in a circle about that edge is less than mg, it won't leave the edge immediately.

Why is the edge the centre of the circle. I was imagining the hoop as the circle so that the centre is at the cog

(edited 10 years ago)

Reply 17

Original post by newblood

Why is the edge the centre of the circle. I was imagining the hoop as the circle so that the centre is at the cog

When the hoop gets to the edge, it starts to rotate about the edge as it starts to fall off the surface. Hence the edge becomes the centre of rotation as long as it now moves on a circular path.

Reply 18