M3 circular motion

Hi guys, your gonna have to bear with me here this might be slighlty long and confusing.

I am looking at question 11/12 vs 14. My question is for question 11/12 do i assume a smooth track and that for it to go around the track in circular motion the speed must be greater than a certain value? So if the speed is less than this value is the car/bicycle unable to go around the track at all? Or will the bicyclist start going round the track but will fall eventually come off the track?

However looking at question 14 i now have doubts in my head. So if the bicyclist in 14 moves at a fast enough speed will she be able to move around the track with no frictional force acting on her, even if the track remains rough? Or does this only work on tracks which are smooth?

Hope it made sense.

Thanks for any input, its quite possible that ive just twisted myself into a confusion for no possible reason😂.

(edited 6 years ago)

Reply 1

Shaanv

OP

17

Original post by Physics Enemy

I don't get the issue. Qs 11&12 state no friction (smooth), Q14 states there's friction (rough). Illogical to say zero friction whilst moving on a rough surface.

They didnt say no friction.

It was worded as 'what speed does the car move if there is no frictional force'. I assumed this meant that there was a frictional force but as the speed was above a certain value the reaction was enough to compensate for friction for some reason.

But now u have cleared it up by confirming that the track in 11/12 is smooth and the one in 14 is not.

I overcomplicated and got muddled.

Thanks appreciate it

Reply 2

ghostwalker

17

Original post by Shaanv

But now u have cleared it up by confirming that the track in 11/12 is smooth and the one in 14 is not.

NONE OF THE TRACKS IN 11/12/14 are SMOOTH!!!!

When they say there is no frictional force (11/12), they mean there is no radial frictional force, and this is because the forces due to motion in a circle, come exactly from the perpendicular reaction; no additional force - due to friction - is required.

Edit: Didn't like the previous wording.

(edited 6 years ago)

Reply 3

Shaanv

OP

17

Original post by ghostwalker

NONE OF THE TRACKS IN 11/12/14 are SMOOTH!!!!

When they say there is no frictional force (11/12), they mean there is no radial frictional force, and this is because the forces due to motion in a circle, exactly counter balance the force down the slope of the bank/track, so no extra force - due to friction - is required to hold the car/bicycle at the given radius.

Sorry ghostwalker, im struggling to understand and visualise why there is no friction involved.

Is it because the centripetal force is equal to R*Sin(theta) so equation 1 reduces down to Fmax cos(theta)=0, therefore Fmax=0 as cos(theta) doesnt equal 0?

(edited 6 years ago)

Reply 4

ghostwalker

17

Original post by Shaanv

Sorry ghostwalker, im struggling to understand and visualise why there is no friction involved.

Is it because the centripetal force is equal to R*Sin(theta) so equation 1 reduces down to Fmax cos(theta)=0, therefore Fmax=0 as cos(theta) doesnt equal 0?

Which question is this relating to? Text / diagram.

Reply 5

Shaanv

OP

17

Original post by ghostwalker

Which question is this relating to? Text / diagram.

It was just a general diagram for circular motion on a banked rough track, i was just trying to figure out in general terms why there is no friction at a certain speed v.

i failed M3

Original post by Shaanv

Sorry ghostwalker, im struggling to understand and visualise why there is no friction involved.

Is it because the centripetal force is equal to R*Sin(theta) so equation 1 reduces down to Fmax cos(theta)=0, therefore Fmax=0 as cos(theta) doesnt equal 0?

IF there is no frictional force, then F=0, and we'd have Rsin theta is the centripetal force.

Consider the situation of a car stationary on a rough banked track:

There will be a frictional force up the bank stopping it from falling down the bank.

Now consider when the car is moving very quickly, the frictional force will be down the bank, to stop the car flying off up the bank.

Finally, consider the car moving at various speeds between the two. If you plot the frictional force against velocity, you'll see that at zero velocity it's a large value postive and at high speed it's a large value negative (up the bank being positive), and it changes continuously as the velocity increases. So, there must be a point, a particular velocity, where the frictional force is zero.

(edited 6 years ago)

Reply 8

Physics Enemy

19

Original post by ghostwalker

NONE OF THE TRACKS IN 11/12/14 are SMOOTH! When they say no frictional force (11/12), they mean there is no radial frictional force ...

Having thought about it tracks can't be smooth, as car's wheels req friction to rotate and move the car linearly, ideally with pure roll. So it must be a rough surface whereby radial friction = 0 at some speed. Thanks for clarifying!

(edited 5 years ago)

Reply 9

Shaanv

OP

17

Original post by ghostwalker

IF there is no frictional force, then F=0, and we'd have Rsin theta is the centripetal force.

Consider the situation of a car stationary on a rough banked track:

There will be a frictional force up the bank stopping it from falling down the bank.

Now consider when the car is moving very quickly, the frictional force will be down the bank, to stop the car flying off up the bank.

Finally, consider the car moving at various speeds between the two. If you plot the frictional force against velocity, you'll see that at zero velocity it's a large value postive and at high speed it's a large value negative (up the bank being positive), and it changes continuously as the velocity increases. So, there must be a point, a particular velocity, where the frictional force is zero.

So just to conclude, there is one speed for which the frictional force equals zero. This depends on radius and angle of slope to horizontal.

If the speed is less than this the friction acts up the slope, if it is greater friction acts down the slope.

So will an object always follow the same path unless its speed is too high or too low for the maximum frictional force to keep the object in circular motion.

Thanks for all the input guys

Reply 10

ghostwalker

17

Original post by Shaanv

So just to conclude, there is one speed for which the frictional force equals zero. This depends on radius and angle of slope to horizontal.

If the speed is less than this the friction acts up the slope, if it is greater friction acts down the slope.

Yes.

So will an object always follow the same path unless its speed is too high or too low for the maximum frictional force to keep the object in circular motion.

Thanks for all the input guys

If you're trying to keep circular motion, then yes.

Edited my initial post, as I didn't like the wording.

M3 circular motion

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