# Block on a turntable

1. Jul 26, 2007

1. The problem statement, all variables and given/known data

The coefficient of friction between a certain brass block and a large revolving turntable is µ = 0.18. How far from the axis of rotation can the block be placed before it slides off the turntable if it is rotating at 33 1/3 rev/min?

2. Relevant equations
how can i do anything with this question without the mass of the block?

3. The attempt at a solution

2. Jul 26, 2007

### G01

You can still set up the equations describing the situation. Try setting up the equations and just use "m" as a placeholder for mass. The fact that you don't know the mass shouldn't be a problem if you have this problem set up right.

Last edited: Jul 26, 2007
3. Jul 26, 2007

how can i get the velocity (to get centrip accel) if i dont know r?
how can i find the right r if i cant get the velocity?

4. Jul 26, 2007

### G01

You may not know v directly, but you do know that the angular fequency is 33 and 1/3 rev/min. Can you use this to find the velocity? You must have some work for this problem. At least show me how you set up the problem or something. I'm really not supposed to give you this much help if you don't show work.

5. Jul 26, 2007

yes i know that v=2pier/T
i have T as 1.8s

But then I need r to get v. \
But the questions is asking me for the optimal r (so it wont slide off).
??

6. Jul 26, 2007

so it seems like a contradiction.
i need r to calcui;llate v, but r is not known

7. Jul 26, 2007

also i realize that F= un
so i will likely do something like equate the un = mv^2/r
and then cncel the n and m somehow to find r

8. Jul 26, 2007

### G01

Your on the right track. What force will the normal force be equal to?

Last edited: Jul 26, 2007
9. Jul 26, 2007

normal will be same as the mass

10. Jul 26, 2007

in the opposing direction i mean

11. Jul 26, 2007

ok so
.18n = mv^2/r
cancel the n and m
.18 = v^2/r

but this is same problem. --i need a value for r in order to calculate v???

12. Jul 26, 2007

### G01

Mass is not a force. I think you mean the normal force will be the same as the weight:

n=mg

.18g=v^2/r

Now, once you know this, you have gotten rid of the m. Now the only problem is that pesky v! If only you had another equation for v in terms of r to plug in there...hint hint.

Last edited: Jul 26, 2007
13. Jul 26, 2007

well this is tangetial vel right
so vt = wr
please tell me what i'm doing wrong?
you can see by now that i know basically whats going on but i'm just tripping over something really stupid.

14. Jul 26, 2007

### G01

You said the formula you need in post #5.

(1)$$v = \frac{2\pi r}{T}$$

Since you have already calculated T, why not use this?

If $$\mu g = \frac{v^2}{r}$$

(2)Then, $$v=\sqrt{r\mu g}$$

So if v = (1) and v = (2)...

Last edited: Jul 26, 2007
15. Jul 26, 2007

yes but what value for r then??

16. Jul 26, 2007