Two identical rollers are mounted with their axes parallel, in a horizontal plane, a distance 2d = 40.4 cm apart. The two rollers are rotating inwardly at the top with the same angular speed. A long uniform board is laid across them in a direction perpendicular to their axes. The board of mass m = 3.52 kg is originally placed so that its center of mass lies a distance x0 = 10 cm from the point midway between the rollers.
The coe�cient of friction between the board and rollers is �k = 0.653. What is the period (s) of the motion?
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To find the period of motion, we need to consider the forces acting on the system and apply Newton's second law of motion. Let's break down the problem step by step:
Step 1: Identify the forces acting on the system.
- Gravity: The weight of the board exerts a downward force.
- Normal force: The normal force is exerted by the rollers on the board, perpendicular to the surface of contact.
- Frictional force: The frictional force acts in the opposite direction to the motion of the board.
Step 2: Set up the equations of motion.
Since the board is rotating around the rollers' axes, we can consider the rotational motion. The torque due to the frictional force provides the restoring torque that creates oscillatory motion. The equation of motion is:
I * α = -k * r
where:
- I is the moment of inertia of the board
- α is the angular acceleration
- k is the coefficient of friction
- r is the distance from the center of mass to the point of contact with the rollers
Step 3: Determine the moment of inertia.
The moment of inertia depends on the shape and mass distribution of the board. Since the board is described as "long uniform," we can assume it is a thin rod rotating about its center. The moment of inertia for a thin rod rotating about its center is:
I = (1/12) * m * L^2
where:
- m is the mass of the board
- L is the length of the board
Step 4: Express the angular acceleration in terms of the variables.
Since the board is rotating inwardly at the top with the same angular speed, the angular acceleration can be expressed as:
α = ω^2 / d
where:
- ω is the angular speed
- d is the distance between the axes of the rollers
Step 5: Substitute the expressions into the equation of motion.
Substituting the expressions for I, α, k, and r into the equation of motion, we get:
(1/12) * m * L^2 * (ω^2 / d) = -k * r
Step 6: Solve for the angular speed ω.
Rearrange the equation to solve for the angular speed ω:
ω^2 = (-k * r * d) / ((1/12) * m * L^2)
ω = sqrt[(-k * r * d) / ((1/12) * m * L^2)]
Step 7: Find the period of motion.
The period of motion is given by:
T = 2π / ω
Substitute the value of ω we found in the previous step into the equation to calculate the period T.