Question

A coin is placed 17.0 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 26.0 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

A coin is placed 17.0 cm from the axis of a rotating turntable of variable speed, the coefficient of static friction between the coin and the turntable is 1.35

Step-by-step explanation:

The coefficient of static friction can be calculated using the given information.

When the coin is at rest, the only force acting on it is the force of static friction.

The force of static friction can be determined using the equation fs = μs * N

Where

fs is the force of static friction

μs is the coefficient of static friction

N is the normal force.

The normal force, N, can be calculated as the product of the coin's mass and the acceleration due to gravity, N = m * g.

For the coin to slide off at a rotational speed of 26.0 rpm, the centripetal acceleration at that speed should exceed the force of static friction, given by a = R * (2πf)²

Where

a is the centripetal acceleration

R is the radius of rotation

f is the frequency of rotation in Hz.

Setting the force of static friction equal to the centripetal acceleration, μs * m * g = R * (2πf)², we can solve for the coefficient of static friction, μs.

Given:

Distance from the axis of rotation (R) = 17.0 cm = 0.17 m

Rotational speed (f) = 26.0 rpm = 26.0 / 60 Hz = 0.433 Hz

Mass of the coin (m) = ?

Acceleration due to gravity (g) = 9.8 m/s²

Using the equation μs * m * g = R * (2πf)², we can solve for μs:

μs = (R * (2πf)²) / (m * g)

Plugging in the given values, we get:

μs = (0.17 * (2π * 0.433)² / (m * 9.8)

So the coefficient of static friction (μs) can be calculated once the mass of the coin (m) is 1.35

Answer 2

Answer: The coefficient of static friction between the coin and the turntable is 0.13.

Step-by-step explanation:

As we know that,

Centripetal force = static frictional force

`(mv^(2))/(r) = F_(s)`

or,
`(mv^(2))/(r) = \mu_(s) * m * g`

v =
`\sqrt{\mu_(s) * r * g}`

or,
`\mu_(s) = (v^(2))/(rg)` ......... (1)

Here, it is given that

r = 17 cm,
`\omega` = 26 rpm,

and v =
`r \omega` ..........(2)

Putting equation (2) in equation (1) we get the following.

`\mu_(s) = (r^(2)\omega^(2))/(rg)`

=
`(17 * 10^(-2) * (26 * [(2 * \pi)/(60)]^(2)))/(9.8)`

= 0.128

= 0.13 (approx)

Thus, we can conclude that the coefficient of static friction between the coin and the turntable is 0.13.