a 5.67g quarter sits on a record player a distance of 5cm from the axis of rotation. if the angular speed of the record player is 33 revolutions per minute, what should the coefficient of friction be to keep it from sliding?
33 *2 pi radians / 60 seconds = 3.46 radians/second = omega
Ac = omega^2 R = 3.46^2 radians^2/seecond^2 * 0.05 meters
= 0.597 meters/s^2
mu m g = m (.597)
mu = .597/g = about .06
To find the coefficient of friction necessary to keep the quarter from sliding on the record player, we can use the equation for centripetal force. The centripetal force is the force required to keep an object moving in a circular path.
The centripetal force can be calculated using the formula:
Fc = m * ω^2 * r
where Fc is the centripetal force, m is the mass of the quarter, ω is the angular velocity (in radians per second), and r is the distance from the axis of rotation.
First, let's convert the angular velocity from revolutions per minute to radians per second:
ω = (33 revolutions / 1 minute) * (2π radians / 1 revolution) * (1 minute / 60 seconds) = 33 * 2π / 60 radians/second
Now, we can plug in the values into the formula:
Fc = (5.67g) * [(33 * 2π / 60)^2] * 0.05
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
To find the coefficient of friction, we need to equate the centripetal force to the product of the coefficient of friction (μ) and the normal force (N). In this case, the normal force acts vertically and is equal to the weight of the quarter, which is given by m * g:
μ * N = Fc
μ * (m * g) = (5.67g) * [(33 * 2π / 60)^2] * 0.05
Simplifying the equation further, we can cancel out the g on both sides:
μ * m = (5.67) * [(33 * 2π / 60)^2] * 0.05
Finally, we can solve for the coefficient of friction (μ) by dividing both sides of the equation by the mass (m):
μ = (5.67) * [(33 * 2π / 60)^2] * 0.05 / m
Now you can calculate the value of the coefficient of friction by plugging in the values of the mass of the quarter (5.67g) and solving the equation.