A penny is placed at the outer edge of a disk (radius = 0.119 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.70 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.
Well, well, well, it seems like we have a penny that wants to experience the thrill of being part of a rotating disk! Let's see how friction can help make that happen.
First, we need to find the minimum coefficient of friction that will keep the penny rotating with the disk. To do that, we need to balance the gravitational force pulling the penny outward with the frictional force trying to keep it in place.
Now, the gravitational force pulling the penny outward is given by the equation F_gravity = m * g, where m is the mass of the penny and g is the acceleration due to gravity.
But wait a minute, a penny's mass is approximately 2.5 grams. However, we're not going to focus on weight here. We're talking about rotational motion! It's time to bring in the centripetal force.
The centripetal force necessary to keep the penny moving in a circle is given by the equation F_centripetal = m * ω^2 * r, where m is the mass, ω is the angular velocity, and r is the radius of the disk.
Since we're dealing with rotations per minute and the period of rotation is given in seconds, we need to convert the period into angular velocity. One complete revolution takes 2π radians, and the period is equal to one complete revolution. So, ω = 2π / T, where T is the period of rotation.
Now that we have all the ingredients, let's mix them up and solve for the minimum coefficient of friction. Here's the equation we need to use:
m * ω^2 * r = μ * m * g
After some clever canceling, we find:
μ = ω^2 * r / g
Use the given values of the radius of the disk (0.119 m) and the period of rotation (1.70 s) to calculate ω. Then plug everything into the equation, and voila! You'll have your answer.
But hey, keep in mind that the minimum coefficient of friction doesn't mean you get to slack off on your cleaning duties. Penny or not, you still gotta keep things spick and span!
To find the minimum coefficient of friction necessary for the penny to rotate along with the disk, we need to consider the centripetal force acting on the penny.
The centripetal force is provided by the frictional force between the penny and the disk's surface.
The formula to calculate centripetal force is:
Fc = m * (v^2 / r)
Where:
- Fc is the centripetal force
- m is the mass of the penny
- v is the velocity of the penny
- r is the radius of the disk
In this case, the penny rotates along with the disk, so the velocity of the penny will be the same as the velocity of any point on the disk's edge.
The formula for the velocity of a point on the edge of a rotating object is:
v = 2 * π * r / T
Where:
- v is the velocity
- r is the radius of the disk
- T is the period of rotation
Substituting this expression for v into the centripetal force formula, we get:
Fc = m * ((2 * π * r / T)^2 / r)
Simplifying, we get:
Fc = 4 * π^2 * m * r / T^2
Now, we can equate the centripetal force Fc to the frictional force Ff:
Ff = μ * m * g
Where:
- Ff is the frictional force
- μ is the coefficient of friction
- m is the mass of the penny
- g is the acceleration due to gravity
Setting Fc = Ff, we can solve for the coefficient of friction:
μ = (4 * π^2 * m * r) / (T^2 * m * g)
Simplifying further, we get:
μ = (4 * π^2 * r) / (T^2 * g)
Now, we can substitute the given values to find the minimum coefficient of friction:
r = 0.119 m
T = 1.70 s
g = 9.8 m/s^2
μ = (4 * π^2 * 0.119 m) / (1.70 s)^2 * 9.8 m/s^2
Calculating this expression, we get:
μ ≈ 0.094
Therefore, the minimum coefficient of friction necessary to allow the penny to rotate along with the disk is approximately 0.094.
To find the minimum coefficient of friction necessary for the penny to rotate along with the disk, we need to analyze the forces acting on the penny.
Let's consider the forces acting on the penny in the rotating frame of reference. There are two forces acting on the penny: the gravitational force (mg), directed vertically downward, and the frictional force (f), directed radially inward towards the center of the disk.
The centripetal force required for an object to move in a circular path is given by Fc = m*(v^2)/r, where m is the mass of the object, v is its velocity, and r is the radius of the circular path.
In this case, since the penny is rotating along with the disk, its velocity is equal to the velocity of any point on the disk's outer edge. The velocity of a point on the edge of a rotating disk is given by v = 2πr/T, where T is the period of rotation.
Therefore, the centripetal force required for the penny to rotate along with the disk is Fc = m*(4π^2r)/T^2.
Now, let's consider the frictional force. The maximum frictional force that can be exerted between two surfaces is given by f_max = μ*N, where μ is the coefficient of friction and N is the normal force.
The normal force acting on the penny is equal to its weight, since the penny is placed on a horizontal surface. Therefore, N = mg.
To ensure that the penny rotates along with the disk, the frictional force (f) should be equal to the centripetal force (Fc). Setting f = Fc, we have μ*N = m*(4π^2r)/T^2.
Substituting N = mg, we get μ*mg = m*(4π^2r)/T^2.
Simplifying, μ = (4π^2r)/(g*T^2).
Now we can calculate the minimum coefficient of friction necessary to allow the penny to rotate along with the disk. Simply substitute the given values for the radius (r = 0.119 m) and the period of rotation (T = 1.70 s) into the equation μ = (4π^2r)/(g*T^2), and calculate the result using the acceleration due to gravity (g = 9.8 m/s^2).
μ = (4π^2 * 0.119) / (9.8 * 1.70^2)
Evaluating this expression gives the minimum coefficient of friction necessary.