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Block on a turntable

  • Thread starter mujadeo
  • Start date
103
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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
 

G01

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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.
 
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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?
Please help
 

G01

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

G01

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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
Your on the right track. What force will the normal force be equal to?
 
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normal will be same as the mass
 
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in the opposing direction i mean
 
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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???
 

G01

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

n=mg

So, your line above should read:

.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.:wink:
 
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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.
 

G01

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You said the formula you need in post #5.

You can plug in v_t=rw for v, but what about this?

(1)[tex]v = \frac{2\pi r}{T}[/tex]

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

If [tex]\mu g = \frac{v^2}{r}[/tex]

(2)Then, [tex]v=\sqrt{r\mu g}[/tex]

So if v = (1) and v = (2)...
 
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yes but what value for r then??
 
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see this is the same problem iv been having all along
they ask whats the optimal r, but to find r i need v, and to find v i need a value for r??
 

G01

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yes but what value for r then??
I have edited my above post to be clearer.

You basically have two equations for v. The trick is to equate the two equations, so you are left with just r as an unknown.
 

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