# Circular motion

1. Jul 14, 2005

### huskydc

The coefficient of friction between a certain brass block and a large revolving turntable is µ = 0.09. How far from the axis of rotation can the block be placed before it slides off the turntable if the turntable rotates at a constant rate of 33 1/3 rev/min (so that it requires time T = 60/33.33 seconds to complete one revolution) ?

V(tangential) = 2pi R / T

i'm not sure if 33.33 rev/min is the V, but i tried plugging in the respective numbers and solve for R, didnt really make much sense.

2. Jul 14, 2005

### Pyrrhus

Draw free body diagram on the block, and use Newton's 2nd Law, also because the velocity vector has constant magnitude it means the tangential acceleration is 0, therefore there's only radial (normal) acceleration. Use this to solve for the radius.

3. Jul 15, 2005

### huskydc

well, i drew a FBD on the block.

and i have :

f = ma
N=mg

from that i solve for a, which = .8829
but i'm still confused as to what that 33.333 rev/min

4. Jul 15, 2005

### Pyrrhus

That's just the speed.

It's simply:

$$F_{friction} = m \frac{v^2}{R}$$

5. Jul 15, 2005

### huskydc

i tried:

F = m (v^2/R)
like you said, V = 33.333

mu g = (v^2 / R) and solve for R this way, didn't work. i've been told that 33.333 is not just V. and if it is, then what's the T for?

6. Jul 15, 2005

### Pyrrhus

You need to have common units!, convert rev/min to m/s!

Last edited: Jul 15, 2005
7. Jul 15, 2005

### huskydc

ok, maybe i'm just dumb with this. (as you can tell physics not my cup of tea)

i did this:

33.333 rev / min x (1 min/60 sec) x (2pi rad/ 1 rev) = 3.49 rad/s

with mu g = (v^2 / R) , R= (v^2)/mu g

am i doing something wrong that i just don't see?

8. Jul 15, 2005

### Pyrrhus

Yes, that's ok. Now i am little confused about the statement so that it requires time T = 60/33.33 seconds to complete one revolution.

Is that how the book states it, it looks to me you would need to use.

$$F_{frictionn} = m \frac{4 \pi^{2} R}{T^{2}}$$

where T is the time for one revolution or the period T, but what you did above, should have worked just the same.

9. Jul 15, 2005

### huskydc

yes, cyclovenom, that one worked. but how did that equation come about?

Thanks so much!!! (physics is so hard....well for me it is...)

10. Jul 15, 2005

### Pyrrhus

Sorry, on the above i meant rev/min to m/s, my bad

it came from

$$v = \frac{2 \pi R}{T}$$

substitute in

$$a_{c} = \frac{v^2}{R} = \frac{4 \pi^{2} R}{T^2}$$