Circular motion, friction and forces

[hankrinsen]

Homework Statement

A sylinder with a mass of 5kg is placed on a distance R from the rotation centre of a carousel. The carousel uses 2 seconds on one round. How far away from the centre can we place the sylinder without it falling of, given a friction number of 0.7 (maximum).

Relevant Equations

friction = μ*n

I am not really sure how to go about this. I have been sick for a couple of weeks and fallen behind a bit. Can anyone help me out please? Thank you

 

[etotheipi]

Hi @hankrinsen, sorry to hear about that! Hope you're better now, and



Have you tried drawing a diagram of the setup?

 

[hankrinsen]

This is attached to the assignment. English is my second language, so i might have misunderstood what you mean by diagram

 

[etotheipi]

That's a good start; what forces are acting on the cylinder, and what are their magnitudes and directions? It is good practice to draw all the forces onto the diagram.

 

[hankrinsen]

I have found that the sum of forces must be facing inwards towards the circle, And I have come to the conclusion that the friction is the centripetal force in this instance. The friction force should be the only force working here. Therefore I am confused as to how i calculate when the cylinder will fall of. Friction force should be 0.7*9.81*5 = 34.335N

 

[wrobel]

I think height of the cylinder is also matter

 

[hankrinsen]

I think height of the cylinder is also matter

How? The height is not given

 

[etotheipi]

I have found that the sum of forces must be facing inwards towards the circle, And I have come to the conclusion that the friction is the centripetal force in this instance. The friction force should be the only force working here. Therefore I am confused as to how i calculate when the cylinder will fall of. Friction force should be 0.7*9.81*5 = 34.335N

That's right, friction provides the centripetal force. The formula ##F = \mu N## you've given corresponds to the maximum possible frictional force (more generally we might say ##F \leq \mu N##).

So you've calculated the maximum centripetal force, and that must correspond to your highest possible radial acceleration. Since the angular velocity is fixed, that naturally corresponds to the highest possible radius.

What is the formula for radial acceleration for circular motion, and then can you apply ##F_r = ma_r##?

 

[wrobel]

How? The height is not given

the cylinder can topple over

 

[hankrinsen]

That's right, friction provides the centripetal force. The formula ##F = \mu N## you've given corresponds to the maximum possible frictional force (more generally we might say ##F \leq \mu N##).

So you've calculated the maximum centripetal force, and that must correspond to your highest possible radial acceleration. Since the angular velocity is fixed, that naturally corresponds to the highest possible radius.

What is the formula for radial acceleration for circular motion, and then can you apply ##F_r = ma_r##?

So this means the highest possible radial acceleration must be 6.867 m/s^2.
The radial acceleration --> A=v^2/r ---> 6.367=v^2/r

To find velocity i find the circumference of one rotation and divide it by time. It takes 2 seconds to finish one rotation with a circumference of 2*pi*r, so the velocity is pi*r

I put it in the equation for max radial acceleration; 6.367 = ((pi*r)^2)/r ---> 6.367 = pi^2*r --> r = 6.367/pi^2

Which gives me that it can be put a maximum og 0.645r from the center without falling off? I might be way off here...

 

[etotheipi]

So this means the highest possible radial acceleration must be 6.867 m/s^2.
The radial acceleration --> A=v^2/r ---> 6.367=v^2/r

To find velocity i find the circumference of one rotation and divide it by time. It takes 2 seconds to finish one rotation with a circumference of 2*pi*r, so the velocity is pi*r

I put it in the equation for max radial acceleration; 6.367 = ((pi*r)^2)/r ---> 6.367 = pi^2*r --> r = 6.367/pi^2

Which gives me that it can be put a maximum og 0.645r from the center without falling off? I might be way off here...

Most of this looks good, though I'm not sure why you included an ##r## in your final answer. You already gave that ##r = \frac{a_r}{{\pi}^2}##, and this should just give you a number. Also, make sure to use the unrounded value for ##a_r##, and try to avoid truncating the decimal too soon!

Notice also that ##a_r = \frac{v^2}{r} = \frac{r^2 \omega^2}{r} = r \omega^{2}##; this last expression can be a little faster if you're dealing with time periods, where ##\omega = \frac{2\pi}{T}##. But your consideration of time for one revolution is also perfectly valid.

In that way, I might have solved it like ##\mu N = mr\omega^2 \implies 0.7 \times 9.81 = r \times (\frac{2\pi}{2})^{2}##.

 

[hankrinsen]

Most of this looks good, though I'm not sure why you included an ##r## in your final answer. You already gave that ##r = \frac{a_r}{{\pi}^2}##, and this should just give you a number. Also, make sure to use the unrounded value for ##a_r##, and try to avoid truncating the decimal too soon!

Notice also that ##a_r = \frac{v^2}{r} = \frac{r^2 \omega^2}{r} = r \omega^{2}##; this last expression can be a little faster if you're dealing with time periods, where ##\omega = \frac{2\pi}{T}##. But your consideration of time for one revolution is also perfectly valid.

In that way, I might have solved it like ##\mu N = mr\omega^2 \implies 0.7 \times 9.81 = r \times (\frac{2\pi}{2})^{2}##.

aha, i thought that since the distance from center is given as r in the assignment, i should give the answer as r aswell, since i can't really know if they are talking about meters or what..

 

[etotheipi]

aha, i thought that since the distance from center is given as r in the assignment, i should give the answer as r aswell, since i can't really know if they are talking about meters or what..

Well in this case ##r## just represents a distance, and you can give a distance as a product of a certain number and a chosen unit. Since you did the calculation with other SI quantities, these units are going to simplify down to metres; however, so long as you're careful, you can give ##r## in whatever units you like.

 

[hankrinsen]

Alright, thank you very much! I understand it a bit more now.