Rather than rotate completely horizontally, a carousel rotates at an angle of 10° from horizontal, with a constant angular velocity of 1.571 RAD/s. A box is placed 1.5 m from the axis of rotation. What’s the minimum coefficient of static friction so the box will not slip?
acceleration down slope = w^2 R cos 10
so m a = m w^2 R cos 10
Force down slope = m g sin 10 + mu m g cos 10
so
g sin 10 + mu g cos 10 = w^2 R cos 10
g tan 10 + mu g = w^2 R
mu = ( w^2 r/g -tan 10)
= 0.201
To find the minimum coefficient of static friction, we need to consider the forces acting on the box placed on the rotating carousel.
First, let's identify the relevant forces:
1. Gravitational force (mg): The weight of the box (mass m multiplied by the acceleration due to gravity g) acts vertically downward.
2. Normal force (N): The reaction force exerted by the carousel on the box acts perpendicular to the surface of contact.
3. Centripetal force (Fc): The force required to keep the box moving in a circle with a constant angular velocity.
4. Frictional force (Ff): The force of static friction between the box and the carousel, preventing slipping.
Now, let's analyze the forces in the horizontal and vertical directions separately.
In the vertical direction:
The weight of the box (mg) is balanced by the normal force (N), so we have:
mg = N
In the horizontal direction:
The only force acting horizontally is the static frictional force (Ff). It provides the necessary centripetal force to keep the box moving in a circle. The centripetal force is given by:
Fc = (m · v^2) / r
Where:
- m is the mass of the box
- v is the linear velocity of the box
- r is the distance from the axis of rotation to the box
To find the linear velocity, we need the angular velocity and the radius:
v = ω · r
Where:
- ω is the angular velocity in radians/s
Substituting this into the centripetal force equation:
Fc = (m · (ω · r)^2) / r
= m · ω^2 · r
Since the static frictional force (Ff) provides the necessary centripetal force (Fc), we have:
Ff = Fc
= m · ω^2 · r
Now, the maximum static frictional force (Fs) can be expressed as:
Fs = μs · N
Where μs is the coefficient of static friction.
To prevent slipping, the maximum static frictional force should be equal to or greater than the required centripetal force:
Fs ≥ Ff
Substituting the expressions for Fs and Ff, we get:
μs · N ≥ m · ω^2 · r
Since N = mg, we can rewrite the inequality as:
μs · mg ≥ m · ω^2 · r
Canceling the mass (m) on both sides, we get:
μs · g ≥ ω^2 · r
Finally, rearranging the equation to solve for the minimum coefficient of static friction (μs), we have:
μs ≥ (ω^2 · r) / g
Now, we can plug in the given values:
- Angular velocity (ω) = 1.571 rad/s
- Radius (r) = 1.5 m
- Acceleration due to gravity (g) = 9.8 m/s^2
Calculating:
μs ≥ (1.571^2 · 1.5) / 9.8
≥ 0.382
Therefore, the minimum coefficient of static friction required to prevent the box from slipping is approximately μs = 0.382.