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?
To find the period of the motion, we need to analyze the forces acting on the system.
First, let's identify the forces acting on the board. We have the gravitational force (mg) acting vertically downward, and the frictional force (f) acting horizontally due to the coefficient of friction between the board and rollers.
Now, let's consider the rotational motion of the board. As the board starts rotating, it experiences a torque due to the frictional force acting on it. This torque causes the board to rotate about its center of mass.
The net torque acting on the board is given by the product of the frictional force (f) and the distance from the center of mass to the axis of rotation (d). In this case, the distance d is half the distance between the rollers, so d = 20.2 cm.
The net torque (τ) can be written as τ = f * d.
Now, let's consider the rotational inertia (I) of the board. For a uniform board rotating about its center of mass, the rotational inertia is given by I = (1/12) * m * L^2, where m is the mass of the board and L is the length of the board.
Given that the mass of the board is m = 3.52 kg and the length of the board is twice the distance between the rollers (2d), we have L = 2 * 20.2 cm = 40.4 cm.
Substituting the values, we have I = (1/12) * 3.52 kg * (40.4 cm)^2.
Now, using Newton's second law for rotational motion, we can write the equation for the angular acceleration (α) as α = τ / I.
Finally, to find the period of motion (T), we can use the formula T = 2π / ω, where ω is the angular speed.
Given that the rollers are rotating with the same angular speed, we can find the angular speed (ω) by considering the linear speed at the top of the roller.
At the top of the roller, the linear speed (v) is equal to the angular speed (ω) multiplied by the distance from the axis of rotation to the top of the roller (20.2 cm).
Using the relationship v = ω * d and rearranging, we have ω = v / d.
Since the linear speed (v) can be found using the equation v = √(g * h), where g is the acceleration due to gravity and h is the height of the board's center of mass above the axis of rotation (in this case, 10 cm), we can substitute the values to find ω.
Once we have the angular speed (ω), we can substitute it into the formula T = 2π / ω to find the period of motion (T).