Question

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.70 and μk= 0.50. The turntable very slowly speeds up.

What is the angular speed in rpm when the coin slides off?

Answer 1

Answer:

$\omega=64.55rpm$

Explanation:

The coin will slide when the centripetal force is too much for the friction to hold, that is, when it is equal to the maximum static friction, so:

$F_{cp}=F_{s}^{max}$

Or:

$ma_{cp}=\mu_sN$

Which means (since on the vertical direction only the normal and weight are acting):

$mr\omega^2=\mu_smg$

So for our values we have:

$\omega=\sqrt{\frac{\mu_sg}{r}}=\sqrt{\frac{(0.7)(9.8m/s^2)}{0.15m}}=6.76rad/s$

Since $1rpm=2\pi rad/60s$, we convert our angular speed using this conversion factor (which is equal to 1 thus not altering our result):

$\omega=6.76rad/s=6.76rad/s(\frac{1rpm}{2\pi rad/60s})=64.55rpm$

Answer 2

We have that for the Question,it can be said that the angular speed in rpm when the coin slides off

N=64.6rpm

From the question we are told

• A 5.0 g coin is placed 15 cm from the center of a turntable.
• The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.70 and μk= 0.50. The turntable very slowly speeds up.  

• What is the angular speed in rpm when the coin slides off?

Generally the equation for the Centripetal acceleration   is mathematically given as

$a_{cp}=rw^2\\\\Therefore\\\\rw^2=Us*g\\\\Therefore\\\\w=\sqrt{\frac{Us*g}{r}}\\\\w=6.8rads$

The Rotation is mathematically given as

$N=\frac{60}{2*pi}$

N=64.6rpm

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