A coin is placed on a large disk which rotates uniformly at a rate of 2 rot/s. The coefficient of friction between the coin and the disk is 0.1. To what distance from the center of the disk should the coin be placed so that the coin may not slip?
I know that mu times g (.1 x 9.8) would go on top, but I don't know what the denominator would be.
the angular speed, or 4pi
To determine the distance from the center of the disk at which the coin should be placed so that it does not slip, we need to consider the centripetal force and the maximum friction force that can be applied without causing slipping.
The centripetal force required to keep the coin moving in a circular path is given by the equation:
Fc = m * R * ω²
Where:
Fc: Centripetal force
m: Mass of the coin
R: Radius of the circular path
ω: Angular velocity of the disk
The maximum friction force that can be applied without causing slipping is given by:
Ff = μ * N
Where:
Ff: Friction force
μ: Coefficient of friction
N: Normal force
The normal force is the force exerted by the disk on the coin, which is equal to the weight of the coin:
N = m * g
Where:
g: Acceleration due to gravity
To prevent slipping, the maximum friction force Ff should be equal to or less than the centripetal force Fc.
μ * N ≤ m * R * ω²
Substituting the expressions for N and ω:
μ * m * g ≤ m * R * (2πf)²
Where:
f: Frequency of rotation, which is equal to ω / (2π)
μ * g ≤ R * (2πf)²
Now we can solve for R. Rearranging the equation:
R ≥ (μ * g) / ((2πf)²)
Substituting the given values:
μ = 0.1
g = 9.8 m/s²
f = 2 rot/s
R ≥ (0.1 * 9.8) / ((2π * 2)²)
Calculating this expression, we get:
R ≥ (0.98) / (25π²)
R ≥ 0.0124 meters (rounded to four decimal places)
Therefore, the coin should be placed at a distance of at least 0.0124 meters from the center of the disk to prevent slipping.
To determine the distance from the center of the disk where the coin should be placed, we need to consider the forces acting on the coin and determine the maximum allowed value for the frictional force.
The first thing to note is that the coin is subject to two forces: the gravitational force pulling it downwards (mg) and the centripetal force (mrω²) due to the rotation of the disk. Here, m represents the mass of the coin, g is the acceleration due to gravity, r is the distance from the center of the disk to the coin, and ω is the angular velocity of the disk.
For the coin not to slip, the maximum value for the frictional force (Ff max) must be equal to or less than the product of the coefficient of friction (μ) and the normal force (Fn).
The normal force (Fn) is equal to the gravitational force acting on the coin (mg) when the system is in equilibrium. Thus, Fn = mg.
To calculate the maximum allowed value for the frictional force, we can substitute the equations for the centripetal force and the normal force into the inequality:
Ff max ≤ μFn
μmg ≥ mrω²
Next, we need to determine the angular velocity (ω) of the disk. Given that the disk rotates at a rate of 2 rotations per second, one full rotation corresponds to 2π radians. Thus, the angular velocity can be calculated as ω = 2π(2).
Finally, we can solve the inequality to find the maximum distance (r) from the center of the disk where the coin should be placed so that it does not slip:
μmg ≥ mrω²
0.1 * m * 9.8 ≥ m * r * (2π(2))²
Simplifying further:
0.1 * 9.8 ≥ r * 16π²
0.98 ≥ r * 16π²
r ≤ 0.98 / (16π²)
Calculating:
r ≤ 0.98 / (16 * π * π)
Using a calculator, we can find:
r ≤ 0.0012 meters (rounded to four decimal places)
Therefore, the coin should be placed no further than 0.0012 meters from the center of the disk.