Circular Motion: Coefficient of Static Friction, u=0.2, Angular Speed, w

In summary, the coefficient of static friction between the slider and the rod is u=0.2 and the inclined rod rotates about a vertical axis AB with a constant angular speed, w, as shown in Figure. The slider is positioned at 0.6 m from B and has an acceleration of 0.52w^2 if it does not slide on the rod. The maximum angular speed the rod can have so that the slider P does not slip up the rod is 4.07 rad/s. At the instant shown, the tangential contact force exerted by the rod on the slider is 5.2 N. The normal force acting on the slider is not mgsin60 because there is a component of acceleration perpendicular to
  • #1
Latios1314
45
0
The coefficient of static friction between the slider and the rod is u=0.2. The inclined rod rotates about a vertical axis AB with a constant angular speed, w, as shown in Figure. At the instant shown, the slider is positioned at 0.6 m from B.
(i) What is the acceleration of slider P if it does not slide on the rod?
(ii) Determine the maximum angular speed the rod can have so that the slider P does not slip up the rod. Draw a free body diagram of the slider showing all the forces acting on it in the vertical plane ABC.
(iii) If the angular speed of the rod increases at a rate of 10 rad/s^2 at the instant shown, before the slider starts to slip, what is the tangential contact force exerted by the rod on the slider?

Note: Answer should be: (i) 0.52w^2 (ii) 4.07 rad/s (iii) 5.2 N

The diagram -
http://pdfcast.org/pdf/circular-motion-3 [Broken]

I understand that we must come up with an equation showing the forces acting along BC and another showing the forces acting along the normal of BC.

But can anyone explain to me why the normal force acting on the slider isn't mgsin60?

My tutor has explained the question in class but he didn't elaborate on this part.
 
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  • #2
why would there be a resultant force acting in the y-direction?
 
  • #3
Latios1314 said:
But can anyone explain to me why the normal force acting on the slider isn't mgsin60?
That would be the case if there were no acceleration, but here there is a component of acceleration perpendicular to the rod.
 
  • #4
Latios1314 said:
why would there be a resultant force acting in the y-direction?
Not sure how you define the y-direction, but the slider is executing circular motion so what direction is it accelerating?
 
  • #5
Doc Al said:
Not sure how you define the y-direction, but the slider is executing circular motion so what direction is it accelerating?

y direction refers to the normal of BC
 
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  • #6
Doc Al said:
That would be the case if there were no acceleration, but here there is a component of acceleration perpendicular to the rod.

but why is there a acceleration perpendicular to the rod? why is normal force greater than mgsin60?
 
  • #7
At first, I've tried solving the equation by equating the focres perpendicular to BC to be n=mgsin60. but can't seem to understand why it doesn't work out.
 
  • #8
Latios1314 said:
but why is there a acceleration perpendicular to the rod?
What's the direction of the acceleration?
 
  • #9
Doc Al said:
What's the direction of the acceleration?

direction meaning?

i was referring to why would there be a resultant force normal to BC.

Shouldn't n=mgsin60? why isn't that the case in this scenario?
 
  • #10
Latios1314 said:
direction meaning?
Acceleration is a vector. Which way is it pointing? (Hint: Circular motion!)

i was referring to why would there be a resultant force normal to BC.
Yes, and to see why there will be a resultant force component in that direction, one first must understand how things are accelerating.

Shouldn't n=mgsin60? why isn't that the case in this scenario?
No, why should it? How did you derive that expression?
 
  • #11
Do you actually mean that the centripetal acceleration acting on the slider is pointed along BC?

Therefore, there are both vertical and horizontal components to the force?
 
  • #12
Does that mean that when we are facing problems regarding circular motion, we must take into account, the force that gave rise to the centripetal force and not the centripetal force itself?
 
  • #13
Latios1314 said:
Do you actually mean that the centripetal acceleration acting on the slider is pointed along BC?
The centripetal acceleration will have components parallel and perpendicular to BC. (What's the direction of that centripetal acceleration?)
Therefore, there are both vertical and horizontal components to the force?
Therefore the force has components parallel and perpendicular to BC.
 
  • #14
Latios1314 said:
Does that mean that when we are facing problems regarding circular motion, we must take into account, the force that gave rise to the centripetal force and not the centripetal force itself?
I don't understand this question. Rephrase it.

When dealing with any sort of problem using Newton's laws, you need to know the acceleration. In the case of uniform circular motion, the acceleration is centripetal.
 
  • #15
Doc Al said:
I don't understand this question. Rephrase it.

When dealing with any sort of problem using Newton's laws, you need to know the acceleration. In the case of uniform circular motion, the acceleration is centripetal.

Not really sure what I'm trying to ask too. Was probably very confused.

Can i say that in this case, the normal force acting on the slider and the friction force acting along BC actually gave rise to the resultant centripetal force pointing towards the centre?
 
  • #16
Latios1314 said:
Can i say that in this case, the normal force acting on the slider and the friction force acting along BC actually gave rise to the resultant centripetal force pointing towards the centre?
Yes. The sum of all forces acting on the slider give a net force pointing towards the center.
 
  • #17
This is not one of the question. But could you explain to me why the slider would fly out if w is above a certain value?

I understand the the tangential velocity of the sider would increase. But would that lead to?
 
  • #18
Latios1314 said:
This is not one of the question. But could you explain to me why the slider would fly out if w is above a certain value?
There is a maximum value of static friction, which is what holds the slider on. Go too fast, and the friction will not be enough to maintain the required centripetal acceleration.
 
  • #19
would need further help.

tangential acceleration=0.6sin60 x 10

why is this so instead of a=06sin60 x 100?
 
  • #20
Latios1314 said:
would need further help.

tangential acceleration=0.6sin60 x 10

why is this so instead of a=06sin60 x 100?

Think I've manage to understand it.

Thanks for the great help. My test is next week. Sure hope i'll be able to do well.
 
  • #21
Got confused again.

Was looking at a question regarding a car traveling on a banked curve.When resolving the vertical forces acting on the car i get n=mgsinθ.

But why doesn't it apply in this case. The free body diagrams end up pretty similar but why isn't n=mgcosθ?
 
  • #22
Latios1314 said:
Got confused again.

Was looking at a question regarding a car traveling on a banked curve.When resolving the vertical forces acting on the car i get n=mgsinθ.

But why doesn't it apply in this case. The free body diagrams end up pretty similar but why isn't n=mgcosθ?
Did you derive it or merely assume it? To derive an expression for the normal force, consider force components and acceleration perpendicular to the surface. Since the acceleration is horizontal and the road is angled, there will be a normal component of acceleration. Apply Newton's 2nd law.
 
  • #23
When a car is traveling on a banked road.

nsinθ=mg

Resolving the horizontal forces ,
ncosθ=(mv^2)/r

is this right?
 
  • #24
Latios1314 said:
When a car is traveling on a banked road.

nsinθ=mg

Resolving the horizontal forces ,
ncosθ=(mv^2)/r

is this right?
How is θ measured? I assume no friction is involved.

If θ is the angle of the road with respect to the horizontal, then you have the sin/cos mixed up.
 
  • #25
yes, indeed. I've got them mixed up.

The angle of the frictionless road is θ to the horizontal.

It should have been

ncosθ = mg
And then by resolving the horizontal forces,
nsinθ = mv^2/r

That's the formula that I've copied down during the lecture. But I don't really understand why do we write it that way instead of

n=mgcosθ

letting n=to a component of the weight of the car.

I don't feel that I've gotten this concept very strongly.
 
  • #26
Latios1314 said:
That's the formula that I've copied down during the lecture. But I don't really understand why do we write it that way instead of

n=mgcosθ
What makes you think this is true? (I suspect you are confusing this result with the situation of something sliding down an incline.)

Do you understand how the formula from class was derived?
 
  • #27
in the case of the banked road, the normal force is greater than the weight of the car?

since nsinθ = mg
 
  • #28
Latios1314 said:
in the case of the banked road, the normal force is greater than the weight of the car?

since nsinθ = mg
That should be ncosθ = mg, but yes, the normal force is greater than the weight. When it goes around the curve, inertia presses it against the road. The road has to push on the car extra hard to keep it going around the curve. (It 'wants' to go straight.)
 
  • #29
I didn't really get the inertia part. what do you mean?
 
  • #30
But back to the case of the inclined rod, can i say that
nsin60=mg?

Tried it. But the answer wasn't right. Could you tell me why?
 
  • #31
Latios1314 said:
But back to the case of the inclined rod, can i say that
nsin60=mg?

Tried it. But the answer wasn't right. Could you tell me why?
Of course it's not right, as I've been saying since my first post. Why do you think it's right? (For my reasons, reread this thread.)
 
  • #32
Reread it.
But i still don't really get it.

There is a resultant centripetal force pointing towards the centre and thus there is a centripetal acceleration pointing towards the centre. The forces along BC is caused by friction and a component of the slider's weight mgcos60.

But how is it different from a car on a banked road?

In the case of a car on a banked road.

ncosθ=mg. It is a component of the normal force ncosθ=mg

but why doesn't a component of the normal force=mg here?
 
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  • #33
Latios1314 said:
But how is it different from a car on a banked road?
It's very similar.
but why doesn't a component of the normal force=mg here?
It does. You are confusing ncosθ = mg with n = mgcosθ. Big difference!
 
  • #34
i mean for the case of a car on a banked road.

ncosθ = mg

But why doesn't nsinθ = mg for the case of the slider?
 
  • #35
Latios1314 said:
i mean for the case of a car on a banked road.

ncosθ = mg

But why doesn't nsinθ = mg for the case of the slider?
Because there is friction. To get an equation for the vertical forces you must include all vertical force components. Friction will have a vertical component.

For the car on a banked road, ncosθ = mg only if there is no friction.
 

1. What is the coefficient of static friction in circular motion with a value of u=0.2?

The coefficient of static friction, represented by the symbol u, is a dimensionless quantity that measures the force required to overcome the static friction between two surfaces. In this case, with a value of u=0.2, it means that it takes 0.2 units of force to overcome the static friction between the two surfaces in circular motion.

2. How is the coefficient of static friction related to the force of friction?

The coefficient of static friction is directly proportional to the force of friction. This means that as the coefficient of static friction increases, so does the force of friction. In other words, the higher the coefficient of static friction, the harder it is to move an object in circular motion due to the force of friction.

3. What is the significance of angular speed in circular motion?

Angular speed, represented by the symbol w, is a measure of how fast an object is rotating in circular motion. It is defined as the change in angular displacement over time. The higher the angular speed, the faster the object is rotating.

4. How is the coefficient of static friction related to the angular speed in circular motion?

The coefficient of static friction is not directly related to the angular speed in circular motion. However, a higher angular speed can result in a higher force of friction, which in turn can increase the coefficient of static friction. This is because a higher angular speed means the object is rotating faster, which can create more friction between the surfaces.

5. How can the coefficient of static friction and angular speed be used to calculate the force of friction in circular motion?

The formula for calculating the force of friction in circular motion is F = u * m * w^2, where F is the force of friction, u is the coefficient of static friction, m is the mass of the object, and w is the angular speed. By plugging in the given values for u and w, the force of friction can be calculated. This can be useful in determining the amount of force needed to overcome the friction and keep the object in circular motion.

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