M2 Normal reaction question

Original post by 000alex

Does (i) mean calculate the normal reaction force?

Yes, they mean the normal reaction force. Since the surface is smooth, the reaction is purely normal.

If so why does the mark scheme say to use Rcos(a)=mg rather than mgcos(a)=R?
Isn't mgcos(a)=R now you usually find the normal reaction force on a slope?

Good question. The answer isn't at all clear to many people who are learning mechanics. They say: "Hmm, there is no acceleration in the direction of the line defined by

R

, since the particle does not leave the surface, hence resolving along the normal to the plane, we have

R=mg\cos \theta

"

Sadly, this is incorrect, and they will get the answer wrong. Why? Well, it is *given* in the question that the particle is executing horizontal circular motion. Hence, you can draw a horizontal acceleration vector on P pointing toward the centre of the circle. This vector can be resolved into a component parallel to

R

and one perp to

R

.

By Newton II, there is thus a net force parallel to

R

, and so

R-mg\cos \theta \ne 0 \Rightarrow R \ne mg \cos \theta

Reply 3

Original post by atsruser

Original post by tiny hobbit

Since the particle is moving in a horizontal circle, which way is the acceleration?

So which way do the forces cancel out?

Ok I get why R can't be mgcos(a).

Can someone help with the math that shows why Rcos(a)=mg? I can't get anything to cancel out

Reply 4

tiny hobbit

Original post by 000alex

Ok I get why R can't be mgcos(a).

Can someone help with the math that shows why Rcos(a)=mg? I can't get anything to cancel out

The only acceleration is horizontal, so upward force = downwards force.

Reply 5

Ok not really needed for this question but how would you work out the total force acting up the plane/slope?
And would this force be equal to mgSin(a) since the particle isn't moving up or down the slope?

Reply 6

Original post by 000alex

Isn't that the reaction force since the particle is in contact with the plane?

Reply 7

Original post by MsFahima

Isn't that the reaction force since the particle is in contact with the plane?

I thought it couldn't be because the reaction is perpendicular to the surface so there is no component of it up or down the plane

Reply 8

Original post by 000alex

I thought it couldn't be because the reaction is perpendicular to the surface so there is no component of it up or down the plane

Hope this helps :smile:

Posted from TSR Mobile

Reply 9

Original post by MsFahima

Hope this helps :smile:

Posted from TSR Mobile

I get the R part of the question :smile:

I'm just wondering what is stopping the particle from moving down the slope?

What would ? be provided by?

55.jpg218.1KB

Reply 10

Original post by 000alex

I get the R part of the question :smile:

I'm just wondering what is stopping the particle from moving down the slope?

What would ? be provided by?

Why do you need to know that? There is no frictional force since it's a smooth plane. Also, since it's moving at a constant angular speed, there is a centripetal force which holds it in place. Right? Correct me if I'm wrong!

(edited 8 years ago)

Reply 11

Original post by MsFahima

I don't need it for this question, I'm just interested in what happens in case a hard question comes up in the exam.

Doesn't the component of the centripetal force act down the plane not up it because the centripetal force is towards the center of the circle??

Reply 12

Original post by 000alex

In the exam, if there is a frictional force they would say "rough plane" or tell you to work out the frictional force so you don't have to worry! No? It's towards the centre... so horizontally.

Reply 13

Original post by MsFahima

In the exam, if there is a frictional force they would say "rough plane" or tell you to work out the frictional force so you don't have to worry! No? It's towards the centre... so horizontally.

Yea but if the plane is smooth something must be keeping it from slipping, and aren't the components of F (Centripetal force) like this?

54.jpg263.3KB

Reply 14

Original post by MsFahima

Hope this helps :smile:

Posted from TSR Mobile

This isn't correct. There is a component of acceleration along the line that you have chosen to resolve.

Reply 15

Original post by 000alex

Yea but if the plane is smooth something must be keeping it from slipping, and aren't the components of F (Centripetal force) like this?

No. Your F is not normal.

There is a component of

mg

tangent to the slope.
There is a component of

mg

normal to the slope.

These forces, together with the normal reaction, cause accelerations tangent and normal to the slope. The resultant of those accelerations (which describes the true motion of the particle) points towards the centre of the circle. This is the centripetal acceleration.

You can't analyse the complete motion of the particle by ignoring some of the accelerations that are acting on it.

Reply 16

Original post by atsruser

This isn't correct. There is a component of acceleration along the line that you have chosen to resolve.

No.. the acceleration is horizontal.

Reply 17

Original post by atsruser

No. Your F is not normal.

There is a component of

mg

tangent to the slope.
There is a component of

mg

So what counters the component of mg tangential to the slope? (so that the particle does not slip)

Is it the component of the centripetal acceleration?
I tried to draw the force due to this centripetal acceleration on the diagram but the tangential component of it points downwards

Reply 18

Original post by MsFahima

Also, since it's moving at a constant angular speed, there is a centripetal force which holds it in place. Right? Correct me if I'm wrong!

This suggests to me that you are not clear on the mechanism by which centripetal force causes uniform circular motion.

The centripetal force causes an acceleration of the particle towards the centre of the circle. Over a short period of time, this acceleration creates a small velocity vector pointing toward the centre of the circle. The resultant of this centripetal velocity and the existing tangential velocity creates a new tangential velocity pointing in a slightly different direction, but of the same magnitude.

It can't be said that the centripetal force holds the particle in position though.

Reply 19