A coin is placed on a record that is rotating at 45 rpm. If the coefficient of static friction between the coin and the record is 0.2, how far from the center of the record can the coin be placed without having it slip off?
.2(mg)=m w^2 r
change 45rpm to rad/sec
solve for r.
To determine how far from the center of the record the coin can be placed without slipping off, we need to consider the balance between the centripetal force and the frictional force.
First, let's determine the centripetal force acting on the coin. The centripetal force required to keep an object of mass m moving in a circle of radius r at a constant speed v can be calculated using the formula:
F = m * (v^2 / r)
In this case, the coin is moving at a constant speed and the radius of the circular motion is the distance from the center of the record to the coin, which we need to find. The velocity can be calculated from the rotational speed of the record:
v = 2 * π * r_p * r
where r_p is the rotational speed in revolutions per minute (rpm), converted to radians per second (rad/s).
Given that r_p = 45 rpm, we have r_p = 45 * 2 * π / 60 = 4.71 rad/s.
Now, we can write the centripetal force equation as:
F = m * (4.71^2 / r)
Next, let's determine the maximum static frictional force that can act between the coin and the record. The maximum static frictional force can be calculated using the equation:
F_friction_max = μ_s * F_normal
where μ_s is the coefficient of static friction and F_normal is the normal force.
The normal force acting on the coin is equal to its weight since it is placed on a level surface. Therefore:
F_normal = m * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now we can write the equation for the maximum static frictional force:
F_friction_max = μ_s * m * g
Since the coin is at the point of slipping, the maximum static frictional force acting between the coin and the record is equal to the centripetal force:
F_friction_max = F = m * (4.71^2 / r)
Setting the maximum static frictional force equal to the centripetal force, we get:
μ_s * m * g = m * (4.71^2 / r)
Simplifying and canceling the mass:
μ_s * g = 4.71^2 / r
Now we can solve for the distance from the center of the record, r:
r = 4.71^2 / (μ_s * g)
Given that the coefficient of static friction μ_s is 0.2 and the acceleration due to gravity g is 9.8 m/s^2, we can calculate the distance:
r = (4.71^2) / (0.2 * 9.8)
r ≈ 11.17 meters
Therefore, the coin can be placed up to approximately 11.17 meters from the center of the record without slipping off.