A coin is placed on an old record turntable which rotates at 78 rpm. The coefficient of friction between the record and the coin is 0.17.
b) At what distance (m) from the center will the coin begin to slide off the record?
c) What is the period of the rotation (s) of the penny?
To determine the distance from the center at which the coin will begin to slide off the record, we need to calculate the critical radius of rotation using the coefficient of friction.
The formula to find the critical radius of rotation is given by:
R = sqrt((μ * g) / (4 * π² * f²))
Where:
R = Critical Radius of Rotation
μ = Coefficient of Friction
g = Acceleration due to gravity (approximately 9.81 m/s²)
f = Frequency of rotation in Hz
Given:
μ = 0.17
f = 78 rpm = 78 / 60 Hz = 1.3 Hz
Plugging in the values into the formula, we can calculate R:
R = sqrt((0.17 * 9.81) / (4 * π² * 1.3²))
Calculating this expression, we find:
R ≈ 0.097 m (rounded to three decimal places)
Therefore, the coin will begin to slide off the record at a distance of approximately 0.097 meters from the center.
To calculate the period of rotation, we can use the formula:
T = 1 / f
Where:
T = Period of Rotation
f = Frequency of rotation in Hz
Given:
f = 1.3 Hz
Plugging in the value, we can calculate T:
T = 1 / 1.3
Calculating this expression, we find:
T ≈ 0.769 s (rounded to three decimal places)
Therefore, the period of rotation of the coin on the record is approximately 0.769 seconds.
To determine the distance from the center at which the coin will begin to slide off the record, you can use the concept of centrifugal force. The centrifugal force is given by the formula:
F_c = m * ω^2 * r
Where F_c is the centrifugal force, m is the mass of the coin, ω is the angular velocity in radians per second, and r is the distance from the center of rotation.
Since the coin is about to slide off the record, the maximum value of the frictional force, F_f, can be defined as:
F_f = μ * F_n
Where μ is the coefficient of friction and F_n is the normal force. In this case, the normal force is equal to the gravitational force acting on the coin, which can be calculated as:
F_n = m * g
Where g is the acceleration due to gravity.
Now, since the coin is at the point of sliding, the frictional force must be equal to the centrifugal force:
F_f = F_c
Substituting the equations above, we have:
μ * F_n = m * ω^2 * r
Simplifying, we get:
μ * m * g = m * ω^2 * r
Canceling out the mass, we have:
μ * g = ω^2 * r
Finally, solving for r, we get:
r = (μ * g) / ω^2
To calculate the period of rotation, we can use the formula:
T = 1 / f
Where T is the period and f is the frequency. The frequency is given by the formula:
f = ω / (2 * π)
Substituting the values, we can solve for the period T.
Now, let's plug in the given values to find the solutions:
b) At what distance (m) from the center will the coin begin to slide off the record?
Given:
ρ = 78 rpm
μ = 0.17
To find ω, we need to convert rpm to radians per second:
ω = (2 * π * ρ) / 60
Substituting the given values, we get:
ω = (2 * π * 78) / 60
Next, we need to substitute the values for g, μ, and ω into the equation for r:
r = (μ * g) / ω^2
c) What is the period of the rotation (s) of the coin?
Given:
ρ = 78 rpm
To find ω, we need to convert rpm to radians per second:
ω = (2 * π * ρ) / 60
Now, we can substitute the values of ω into the equation for period:
T = 1 / f
That's how you can calculate the distance at which the coin will begin to slide off the record and the period of rotation of the coin.