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?
Well, well, well, let's roll with this question! To find the period of motion, we need to consider the balance between the gravitational force and the frictional force acting on the board. Let's get clowning around with the calculations:
The gravitational force acting on the board can be found using the formula Fg = mg, where m is the mass of the board and g is the acceleration due to gravity. Plugging in the values, we have Fg = 3.52 kg * 9.8 m/s^2.
Now, the frictional force acting on the board can be found using the formula Ff = μk * N, where μk is the coefficient of friction and N is the normal force. Since the board is at rest, the normal force is equal to the gravitational force, N = Fg. Therefore, Ff = μk * Fg.
Since the board is in rotational equilibrium, the frictional force must provide the centripetal force needed to keep the board in circular motion. The centripetal force can be found using the formula Fc = (m * x0 * w^2) / d, where m is the mass of the board, x0 is the distance from the point midway between the rollers to the center of mass of the board, w is the angular speed of the rollers, and d is the distance between the rollers.
Equating the frictional force and the centripetal force, we have μk * Fg = (m * x0 * w^2) / d.
Now we can solve for the angular speed w by rearranging the equation:
w^2 = (μk * Fg * d) / (m * x0)
Once we have the angular speed w, we can find the period T using the formula T = 2π / w.
So, let's put on our clown noses and plug in the numbers to find the period! May the laughter be with you!
Fg = 3.52 kg * 9.8 m/s^2
Ff = 0.653 * Fg
w^2 = (0.653 * Fg * d) / (m * x0)
T = 2π / w
Now, someone make me a balloon animal while I do the math!
To find the period of the motion, we need to consider the forces acting on the board and determine when it completes a full oscillation.
1. Calculate the gravitational force acting on the board:
F_gravity = m * g
where m = mass of the board = 3.52 kg
g = acceleration due to gravity ≈ 9.8 m/s^2
2. Calculate the frictional force acting on the board:
F_friction = μk * N
where μk = coefficient of friction = 0.653
N = normal force
The normal force N can be determined by balancing the vertical forces:
N = F_gravity
3. Calculate the net torque acting on the board:
The net torque is the torque due to the gravitational force and the torque due to the frictional force.
τ_net = τ_gravity + τ_friction
τ_gravity = F_gravity * r
where r = x0 (the distance of the board's center of mass from the point midway between the rollers)
τ_friction = F_friction * d (distance between the rollers)
4. Equate the net torque to the moment of inertia multiplied by the angular acceleration:
τ_net = I * α
The moment of inertia of the board can be approximated as:
I = (1/3) * m * L^2
where L is the length of the board (the distance between the rollers).
The angular acceleration is related to the period T by the equation:
α = (2π) / T
5. Substituting the torque equation and simplifying:
τ_net = I * α
(F_gravity * r) + (F_friction * d) = [(1/3) * m * L^2] * [(2π) / T]
(m * g * r) + (μk * N * d) = [(1/3) * m * L^2] * [(2π) / T]
Substituting N = F_gravity:
(m * g * r) + (μk * m * g * d) = [(1/3) * m * L^2] * [(2π) / T]
Simplifying and isolating T:
T = [(1/3) * L^2 * (2π)] / [(g * r) + (μk * g * d)]
6. Substituting the given values:
L = 2d = 40.4 cm = 0.404 m
r = x0 = 10 cm = 0.1 m
μk = 0.653
g ≈ 9.8 m/s^2
T = [(1/3) * (0.404^2) * (2π)] / [(9.8 * 0.1) + (0.653 * 9.8 * 0.404)]
T ≈ 0.8701 s
Therefore, the period of motion is approximately 0.8701 seconds.
To find the period of the motion, we need to first analyze the forces acting on the board and use the equations of rotational motion.
1. The gravitational force acting on the board can be split into two components: one perpendicular to the board's surface and one parallel to the board's surface.
2. The perpendicular component of the gravitational force does not affect the rotation of the board. Hence, we will only consider the parallel component. The parallel component of the gravitational force is given by F_grav = m * g * sin(theta), where m is the mass of the board, g is the acceleration due to gravity, and theta is the angle between the board and the horizontal.
3. The friction force between the rollers and the board provides the torque needed to rotate the board. The friction force can be calculated using the equation F_friction = mu_k * F_norm, where mu_k is the coefficient of friction between the board and the rollers, and F_norm is the normal force acting on the board. In this case, the normal force is equal to the perpendicular component of the gravitational force, which is F_grav = m * g * sin(theta).
4. The torque acting on the board is given by the equation τ = F_friction * d, where d is the distance from the center of mass of the board to the point of rotation (distance between the rollers).
5. Using Newton's second law for rotational motion, τ = I * alpha, where I is the moment of inertia of the board and alpha is the angular acceleration. For a long uniform board rotating about an axis perpendicular to its length, the moment of inertia is I = (1/3) * m * L^2, where L is the length of the board.
6. The angular acceleration alpha can be related to the angular speed omega and the period T through the equation alpha = (2 * pi) / T.
7. Combining these equations and solving for the period T, we can find the answer.
Please note that theta, the angle between the board and horizontal, is not given in the question. To complete the calculation, you would need to provide the value of theta.