An object of mass 1.4kg is placed 11.5 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the object remains fixed on the turntable until a rate of 37 rpm is reached, at which point the object slides off. What is the coefficient of static friction between the object and the turntable?
To determine the coefficient of static friction between the object and the turntable, we need to analyze the forces acting on the object and consider the condition for impending motion (when the object is just about to slide off).
First, let's consider the forces acting on the object. We have the gravitational force (mg) acting vertically downward and the static friction force (F_friction) acting horizontally inward towards the axis of rotation.
For the object to remain fixed on the turntable, the static friction force must provide the necessary centripetal force to keep the object moving in a circle. At the point just before sliding off, the centripetal force is given by:
F_centripetal = m * (ω^2) * R
Where:
- F_centripetal is the centripetal force,
- m is the mass of the object (1.4 kg),
- ω is the angular velocity of the turntable (in rad/s), and
- R is the distance from the axis of rotation to the object (11.5 cm or 0.115 m).
Since we are given the speed of the turntable in terms of revolutions per minute (rpm), we need to convert it to rad/s using the following conversion:
ω = (2π * n) / 60
Where:
- ω is the angular velocity in radians per second,
- n is the speed in revolutions per minute.
Converting the given speed of 37 rpm to rad/s:
ω = (2π * 37) / 60 ≈ 3.878 rad/s
Now we have all the necessary information to calculate the centripetal force:
F_centripetal = m * (ω^2) * R
= 1.4 kg * (3.878 rad/s)^2 * 0.115 m
≈ 0.755 N
Since the object is just about to slide off the turntable, the maximum static friction force (F_friction) is equal to the centripetal force:
F_friction = F_centripetal
= 0.755 N
Finally, we can determine the coefficient of static friction (μ_static) using the equation:
F_friction = μ_static * N
Where N is the normal force, which is equal to the weight of the object (mg).
Substituting the values, we get:
0.755 N = μ_static * (1.4 kg * 9.8 m/s^2)
Solving for μ_static:
μ_static = 0.755 N / (1.4 kg * 9.8 m/s^2)
≈ 0.058
Therefore, the coefficient of static friction between the object and the turntable is approximately 0.058.