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Aiyume · Oct 7, 2010, 09:28 PM

A small block sits 0.15 m from the center of a horizontal turntable whose frequency of rotation can be smoothly increased. If the coefficient of static friction between the block and the turntable is 0.55, at what frequency will the block begin to slide off? (Draw a force diagram).

f_s = U_s F_n
-f_s = ma
a = -f_s/m = -(U_s mg)/m = -U_s g
a = v^2/R
-U_s g = v^2/R
v = Sq rt(-R u_s g)

I am not sure what the question is really asking. What's frequency? Is it velocity?

About the diagram:
(a) A circle with radius 0.15m, centripetal force point toward the centre, velocity is a tangent line on the circle.
(b) Another diagram, with the Mass in the middle, F_n pointing upward, Fg pointing downward, F (and acceleration) pointing to the left, f_s pointing to the right.

Unknown008 · Oct 8, 2010, 01:16 AM

Frequency is the inverse of periodic time, the time for the turntable to complete one complete rotation.

In this problem, it is the frictional force which provides the centripetal force. When the Frictional force becomes too little, it cannot provide enough centripetal force for the block to stay on the turntable.

So, at critical point (when the block is about to slide);

$F_s = F_c$

$\mu F_n = \frac{mv^2}{r}$

Now, $v = r\omega = r$\frac{2\pi}{T}$$

And; $f = \frac{1}{T}$

So, combining all those, we get:

$\mu mg= \frac{m(2\pi rf)^2}{r}$

Solve for f :)

Aiyume · Oct 8, 2010, 08:45 AM

Wow... Thank you so much for your help!
Do you know any web site that has physics questions that I can practice?

Unknown008 · Oct 8, 2010, 08:53 AM

No, sorry :(

Aiyume · Oct 9, 2010, 10:46 AM

Don't be sorry~ ^^~