A penny is placed at the outer edge of a disk (radius = 0.159 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.50 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk
To find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk, we need to consider the forces acting on the penny.
1. Centripetal force: The penny will experience a centripetal force that keeps it moving in a circular path. This force is given by the equation:
F = (m * v^2) / r
where m is the mass of the penny, v is its tangential velocity, and r is the radius of the disk.
2. Frictional force: The minimum coefficient of friction necessary to allow the penny to rotate along with the disk can be determined by considering the maximum frictional force that can act on the penny without causing it to slip. This maximum frictional force can be expressed as:
f_max = μ * N
where μ is the coefficient of friction and N is the normal force acting on the penny. The normal force is equal to the weight of the penny.
3. Tangential velocity: The tangential velocity of the penny can be found by dividing the circumference of the circle it travels along by the period of rotation:
v = 2πr / T
where T is the period of rotation.
To find the minimum coefficient of friction, we can equate the centripetal force to the maximum frictional force:
(m * v^2) / r = μ * N
Now, let's substitute the equations for centripetal force, tangential velocity, and normal force:
(m * (2πr / T)^2) / r = μ * (m * g)
where g is the acceleration due to gravity.
Simplifying the equation:
(4π^2 * r) / T^2 = μ * g
Finally, solve for the coefficient of friction:
μ = (4π^2 * r) / (T^2 * g)
Substituting the given values:
μ = (4 * 3.14^2 * 0.159 m) / (1.50 s)^2 * 9.8 m/s^2
Calculating the value:
μ ≈ 0.35
Therefore, the minimum coefficient of friction necessary to allow the penny to rotate along with the disk is approximately 0.35.
To find the minimum coefficient of friction necessary for the penny to rotate along with the disk, we need to consider the forces acting on the penny.
First, let's analyze the forces acting on the penny:
1. The force of gravity acts vertically downward on the penny. We can denote this force as mg, where m is the mass of the penny and g is the acceleration due to gravity.
2. The normal force acts perpendicular to the surface of the disk and opposes the force of gravity. Denoted as N, it is equal in magnitude but opposite in direction to mg.
Since the penny is placed at the outer edge of the disk and the disk rotates, the penny experiences a centripetal force that tries to make it move towards the center of the rotation. This force is provided by the friction between the penny and the disk surface.
The centripetal force is given by the equation Fc = m * ac, where m is the mass of the penny and ac is the centripetal acceleration. In this case, the centripetal force is the frictional force between the penny and the disk.
The centripetal acceleration, ac, can be calculated using the formula ac = (2 * pi * r) / T, where r is the radius of the disk and T is the period of rotation.
By equating the centripetal force with the frictional force, we can solve for the minimum coefficient of friction, μ, needed for the penny to rotate along with the disk:
μ * N = m * ac
We need to solve for μ, so we need to express N in terms of the mass of the penny and acceleration due to gravity:
N = mg
Substituting the value of ac:
μ * mg = m * (2 * pi * r) / T
Simplifying and canceling out the mass of the penny:
μ * g = (2 * pi * r) / T
Finally, solving for the minimum coefficient of friction:
μ = (2 * pi * r) / (T * g)
Let's substitute the given values into the formula to get the minimum coefficient of friction required for the penny to rotate along with the disk.