# Thread: Stuck on the concept of a particle moving on a horizontal circle :help:

1. ## Stuck on the concept of a particle moving on a horizontal circle :help:

Hello, I am struggling to imagine what is going on here. My questions are:
1. What does it mean that the particle is on the point of slipping? If the particle slips, in which direction does it move?
2. What force is moving the disc itself? Would the disk stop moving if the particle is not present there? If not, what does the sentence mean: "Here, the centripetal force is provided by the frictional force F N between the particle and the disc."
3. If it was a smooth horizontal disc rather than rough horizontal disc, what would happen?

2. 1. I think that this means that it is currently in equalibrium, ie not moving at the moment.

3. Smooth disc would equal no friction, so remove the friction from you equation and see what happens, could mean it moves up/down the disc?

Sorry I can't answer 2, have had no experience with centripetal forces, from what I can see question 1 and 3 are just basic mechanics.

Hope this helps a bit

Craig

3. Originally Posted by craig
1. I think that this means that it is currently in equalibrium, ie not moving at the moment.

3. Smooth disc would equal no friction, so remove the friction from you equation and see what happens, could mean it moves up/down the disc?

Sorry I can't answer 2, have had no experience with centripetal forces, from what I can see question 1 and 3 are just basic mechanics.

Hope this helps a bit

Craig
Actually I am confused as to the role of the friction here. What does it do and what happens if it is absent? And if I do not understand the role of friction, how am i supposed to understand the sum?

Hello, I am struggling to imagine what is going on here. My questions are:
1. What does it mean that the particle is on the point of slipping? If the particle slips, in which direction does it move?
2. What force is moving the disc itself? Would the disk stop moving if the particle is not present there? If not, what does the sentence mean: "Here, the centripetal force is provided by the frictional force F N between the particle and the disc."
3. If it was a smooth horizontal disc rather than rough horizontal disc, what would happen?

I can't see your attachment. Nevertheless it sounds like you have particle sitting on a turntable. Relative to an external observer:

1. With no friction the particle will not move (the turntable slips underneath the particle).

2. With friction:

$\displaystyle F_{net} = \text{Friction force} = \frac{mv^2}{r}$ .... (1)

(i) If the particle is not on the point of slipping it moves in a circle of radius r specified by equation (1).

(ii) If the particle is on the point of slipping then

$\displaystyle \text{Friction force} = \mu R$ .... (2)

The particle moves in a circle specified by equations (1) and (2).

(iii) If the particle slips, there is still friction. The particle moves radially outwards in a spiral.

5. place an object on a turntable ... spin the turntable.

the object follows a circular path because the force of static friction provides the necessary centripetal force for the object to move in a circular path.

the centripetal force is directed toward the center and, once again, is provided by the force of static friction. but static friction has a maximum value, $\displaystyle f_{s \, max} = \mu_s \cdot N$. The max force of static friction will be reached if the turntable spins fast enough ... that maximum angular speed can be determined by setting the max force of static friction equal to the centripetal force ...

$\displaystyle \mu_s \cdot N = F_c$

$\displaystyle \mu_s \cdot N = mr\omega^2$

since the object sits flat on the turntable, $\displaystyle N = mg$ ...

$\displaystyle \mu_s \cdot mg = mr\omega^2$

$\displaystyle \mu_s \cdot g = r\omega^2$

$\displaystyle \sqrt{\frac{\mu_s g}{r}} = \omega$

if $\displaystyle \omega$ reaches the value calculated above, the object will start to slip and will not maintain its circular motion ... it will obey Newton's 1st law and will try to maintain a speed in a straight line as much as kinetic friction will allow it.

6. Originally Posted by mr fantastic
I can't see your attachment. Nevertheless it sounds like you have particle sitting on a turntable. Relative to an external observer:

1. With no friction the particle will not move (the turntable slips underneath the particle).

2. With friction:

$\displaystyle F_{net} = \text{Friction force} = \frac{mv^2}{r}$ .... (1)

(i) If the particle is not on the point of slipping it moves in a circle of radius r specified by equation (1).

(ii) If the particle is on the point of slipping then

$\displaystyle \text{Friction force} = \mu R$ .... (2)

The particle moves in a circle specified by equations (1) and (2).

(iii) If the particle slips, there is still friction. The particle moves radially outwards in a spiral.
I finally understand. Thanks.
If you click the image you may see it. My bad, I uploaded it somewhere where it requires password, but you being so intuitive I have the problem solved

7. Originally Posted by mr fantastic
I can't see your attachment. Nevertheless it sounds like you have particle sitting on a turntable. Relative to an external observer:

1. With no friction the particle will not move (the turntable slips underneath the particle).

2. With friction:

$\displaystyle F_{net} = \text{Friction force} = \frac{mv^2}{r}$ .... (1)

(i) If the particle is not on the point of slipping it moves in a circle of radius r specified by equation (1).

(ii) If the particle is on the point of slipping then

$\displaystyle \text{Friction force} = \mu R$ .... (2)

The particle moves in a circle specified by equations (1) and (2).

(iii) If the particle slips, there is still friction. The particle moves radially outwards in a spiral.