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 motion, we first need to understand the forces acting on the board and how they affect its motion.
Since the rollers are rotating inwardly at the top with the same angular speed, they will exert a force on the board due to friction. This frictional force will act opposite to the motion of the board, trying to prevent it from sliding off the rollers.
The normal force between the board and the rollers will act towards the center of the rollers, perpendicular to the board's surface.
Considering the rotational motion of the board and the forces acting on it, we can analyze it as a torque problem.
Let's break down the steps to find the period of motion:
1. Calculate the gravitational force acting on the board:
The gravitational force can be calculated by multiplying the mass of the board (m = 3.52 kg) by the acceleration due to gravity (g = 9.8 m/s²).
F_gravity = m * g
2. Calculate the frictional force:
The frictional force can be calculated by multiplying the coefficient of friction (µk = 0.653) by the normal force acting on the board.
F_friction = µk * F_normal
3. Calculate the normal force:
The normal force can be calculated by summing up the gravitational force and the component of the normal force due to the acceleration of the board.
F_normal = F_gravity + m * a
Here, a represents the acceleration of the board.
4. Calculate the net torque acting on the board:
The net torque is the difference between the torque due to the frictional force and the torque due to the gravitational force.
τ_net = τ_friction - τ_gravity
The torque due to the frictional force can be calculated by multiplying the frictional force by the distance of the board's center of mass (x0) from the midpoint between the rollers.
τ_friction = F_friction * x0
The torque due to the gravitational force can be calculated by multiplying the gravitational force by the distance of the board's center of mass (x0) from the midpoint between the rollers.
τ_gravity = F_gravity * x0
5. Calculate the moment of inertia of the board:
The moment of inertia of the board can be calculated using the formula for a uniform board rotating about its center.
I = (1/12) * m * L²
Here, L represents the length of the board.
6. Calculate the angular acceleration:
The angular acceleration (α) can be calculated by dividing the net torque by the moment of inertia.
α = τ_net / I
7. Calculate the period:
The period (T) can be calculated using the formula:
T = (2π) / √α
Here, √α represents the square root of the angular acceleration.
By following these steps and plugging in the given values, you can find the period (T) of the motion.
To determine the period of the motion, we need to find the time it takes for the board to complete one full cycle.
First, let's find the net force acting on the board. The only external force acting on the board is the friction force between the board and the rollers.
The friction force can be calculated using the equation:
Friction force (Ff) = coefficient of friction (μk) * normal force (N)
The normal force is equal to the weight of the board since it is in equilibrium:
Normal force (N) = mass (m) * gravitational acceleration (g)
Substituting the given values:
Normal force (N) = 3.52 kg * 9.8 m/s^2
Normal force (N) = 34.496 N
Now, calculate the friction force:
Friction force (Ff) = 0.653 * 34.496 N
Friction force (Ff) = 22.542688 N
Since the two rollers are rotating inwards at the top with the same angular speed, the board experiences a centripetal force due to the rotation. This force can be calculated using the equation:
Centripetal force (Fc) = mass (m) * angular velocity (ω)^2 * distance from the center of mass to the point of rotation (x0)
The angular velocity (ω) can be found by using the relationship between linear velocity (v) and angular velocity (ω):
v = ω * r
where r is the radius of the roller, given as 20.2 cm.
The linear velocity (v) can be found using the formula:
v = 2πr / T
where T is the period of motion we want to find.
Substituting the given values:
v = 2π * 0.202 m / T
Rearranging the equation gives:
ω = v / r = 2π * 0.202 m / T * 0.202 m
Now, calculate the centripetal force:
Centripetal force (Fc) = 3.52 kg * (2π * 0.202 m / T * 0.202 m)^2 * 0.1 m
Since the board is in equilibrium, the net force acting on it is zero. Therefore, the centripetal force and the friction force must be equal:
Friction force (Ff) = Centripetal force (Fc)
22.542688 N = 3.52 kg * (2π * 0.202 m / T * 0.202 m)^2 * 0.1 m
Solving for T:
T = 2π / sqrt((22.542688 N / 3.52 kg / 0.1 m) * (0.202 m)^2)
Calculating this value will give you the period (T) of the motion.