A penny is placed on a disk that is rotating at 2 revolutions per second. The coefficient of friction between the penny and the disk is 0.4. What is the maximum radius at which the penny will stay on the disk?
a. 2.5cm
b. 100cm
c. 1cm
d. 10cm
e. 25cm
mu m g = m v^2/r = m omega^2 r
mu g = omega^2 r
omega = 2 *2pi radians/s = 4 pi rad/s
mu = .4
g = 9.81
so
r = .4 * 9.81 /16 pi^2
= .0248 meters = 2.5 cm
To determine the maximum radius at which the penny will stay on the disk, we can use the concept of centripetal force.
The centripetal force required to keep the penny moving in a circle is provided by the force of friction between the penny and the rotating disk. This force of friction can be calculated using the equation:
frictional force = coefficient of friction * normal force
In this case, the normal force is the weight of the penny, which can be calculated using the equation:
weight = mass * gravity
Assuming the mass of the penny is negligible compared to the force exerted by gravity, we can simplify the equation further:
weight = mass * gravity = mg
Now, the centripetal force can be calculated using the equation:
centripetal force = mass * (2πr / T)^2
where r is the radius of the disk, T is the time for one revolution (1 / 2 revolutions per second), and π is approximately 3.14159.
Equating the centripetal force to the frictional force, we have:
mass * (2πr / T)^2 = coefficient of friction * mass * gravity
Simplifying the equation further, we get:
(2πr / T)^2 = coefficient of friction * gravity
Now, let's plug in the given values:
coefficient of friction = 0.4
gravity = 9.8 m/s^2
T = 1 / (2 revolutions per second) = 1 / (2 * 2) = 1 / 4 = 0.25 seconds
Substituting these values into the equation, we have:
(2πr / 0.25)^2 = 0.4 * 9.8
Now, let's solve for r:
(2πr / 0.25)^2 = 3.92
Taking the square root of both sides, we have:
2πr / 0.25 = √(3.92)
Simplifying further:
2πr = 0.25 * √(3.92)
Dividing by 2π, we find:
r = (0.25 * √(3.92)) / (2π)
Using a calculator, we can find the approximate value of r to be 0.138 meters or 13.8 centimeters.
Therefore, the maximum radius at which the penny will stay on the disk is approximately 13.8 centimeters.
Out of the given choices, the closest option is d. 10cm.
To find the maximum radius at which the penny will stay on the disk, we need to consider the centripetal force required to keep the penny moving in a circular path. In this case, the centripetal force is provided by the frictional force between the penny and the disk.
We can start by finding the centripetal force required using the formula:
F = m * a
Where:
F is the centripetal force
m is the mass of the penny
a is the centripetal acceleration
The penny is in circular motion, so the centripetal acceleration can be calculated as:
a = r * ω^2
Where:
r is the radius of the circular path
ω is the angular velocity of the disk
Given that the disk is rotating at 2 revolutions per second, we can calculate the angular velocity:
ω = 2π * f = 2π * 2 = 4π rad/s
Next, we can calculate the centripetal force required:
F = m * r * ω^2
The frictional force between the penny and the disk provides the centripetal force, so we can write:
F = μ * m * g
Where:
μ is the coefficient of friction between the penny and the disk
g is the acceleration due to gravity
Now we can equate the two expressions for centripetal force:
μ * m * g = m * r * ω^2
The mass of the penny cancels out, giving us:
μ * g = r * ω^2
Finally, we can solve for r:
r = (μ * g) / ω^2
Substituting the given values for the coefficient of friction (μ = 0.4), acceleration due to gravity (g = 9.8 m/s^2), and angular velocity (ω = 4π rad/s), we can calculate the maximum radius at which the penny will stay on the disk:
r = (0.4 * 9.8) / (4π)^2
r ≈ 0.992 m
The radius is approximately 0.992 m, which is equivalent to 99.2 cm.
Therefore, the correct answer is d. 10cm.