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=0.80 and μk=0.50. The turntable very slowly speeds up to 60 rpm. Does the coin slide off?

Answer 1

Answer:

the coin does not slide off

Explanation:

mass (m) = 5 g = 0.005 kg

distance (r) = 15 cm = 0.15 m

static coefficient of friction (μs) = 0.8

kinetic coefficient of friction (μk) = 0.5

speed (f) = 60 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the angular speed of the table

ω = 2πf

ω = 2 x π x 60 x $\frac{1}{60}$

ω  = 6.3 s^{-1]

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force (Fs) = ma

static force (Fs) = μs x Fn = μs x m x g

Fs = 0.8 x 0.005 x 9.8 = 0.0392 N

Fs = ma

0.0392 = 0.005 x a

a = 7.84 m/s^{2}

$(Vmax)^{2}$ = a x r

$(Vmax)^{2}$ = 7.84 x 0.15

Vmax = 1.08 m/s

ωmax = $\frac{Vmax}{r}$

ωmax = $\frac{1.08}{0.15}$ = 7.2 s^{-1}

now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm

Fmax = $\frac{ωmax}{2π}$

Fmax = $\frac{7.2}{2 x π}$ = 68.7 rpm

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.