Question

A coin is placed 13.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 38.0 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

Answer 1

Answer:

μs = 0.2098

Explanation:

Given

R = 13.0 cm = 0.13 m

ω = 38 r.p.m = (38 rev/min)(2π/1 rev)(1 min/60 s) = 3.98 rad/s

If we apply Newton's Second Law

∑ Fc = m*ac

⇒ Ff = m*ac

where Ff is the force of friction:

Ff = μs*N = μs*(m*g) ⇒ Ff = μs*m*g

and

ac = ω²*R

then

μs*m*g = m*(ω²*R)   ⇒    μs = ω²*R / g

⇒   μs = (3.98 rad/s)²*(0.13 m) / (9.81 m/s²)

⇒   μs = 0.2098

Answer 2

The  coefficient of static friction between the coin and the turntable is μs = 0.2098

Calculation of the  coefficient of static friction:

Since

R = 13.0 cm = 0.13 m

ω = 38 r.p.m = (38 rev/min)(2π/1 rev)(1 min/60 s) = 3.98 rad/s

Now here we applied Newton's Second Law

So,

∑ Fc = m*ac

⇒ Ff = m*ac

Here Ff is the force of friction:

Now

Ff = μs*N = μs*(m*g) ⇒ Ff = μs*m*g

And

ac = ω²*R

So,

μs*m*g = m*(ω²*R)   ⇒    μs = ω²*R / g

μs = (3.98 rad/s)²*(0.13 m) / (9.81 m/s²)

μs = 0.2098

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