A bar of mass and negligible height is lying horizontally across and perpendicular to a pair of counter rotating rollers as shown in the figure. The rollers are separated by a distance . There is a coefficient of kinetic friction between each roller and the bar. Assume that the bar remains horizontal and never comes off the rollers, and that its speed is always less than the surface speed of the rollers. Take the acceleration due to gravity to be .
(a) Find the normal forces and exerted by the left and right rollers on the bar when the center of the bar is displaced a distance from the position midway between the rollers. Express your answers in terms of , , and (enter m for , x for , d for and g for ).
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(b) Find the differential equation governing the horizontal displacement of the bar . Express your answer in terms of , , and (enter x for , d for , mu_k for and g for ).
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(c) The bar is released from rest at at . Find the subsequent location of the center of the bar, . Express your answer in terms of , , , and (enter x_0 for , d for , mu_k for , t for and g for ).
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To answer the given questions, we need to apply the principles of mechanics, specifically Newton's laws of motion and the concept of friction.
(a) To calculate the normal forces exerted by the left and right rollers on the bar, we need to consider the forces acting on the bar at the displaced position. The forces acting on the bar are its weight (mg) acting downwards and the frictional forces (f) acting horizontally.
Let's assume the left roller is on the left side of the bar and the right roller is on the right side of the bar. At the displaced position, the left side of the bar is tilted upward due to the displacement. We can calculate the normal force at the left roller (Nl) using the following equation:
Nl - mg = 0 (since the bar is not lifting off the rollers)
Nl = mg
Similarly, at the right roller, the right side of the bar is tilted downward due to the displacement. We can calculate the normal force at the right roller (Nr) using the following equation:
Nr + mg = 0 (since the bar is not lifting off the rollers)
Nr = -mg
So, the normal forces exerted by the left and right rollers on the bar are Nl = mg and Nr = -mg, respectively.
(b) To find the differential equation governing the horizontal displacement of the bar, we need to consider the net force acting on the bar in the horizontal direction. The net force is the difference between the frictional force on the left side (-f) and the frictional force on the right side (f). Since the bar remains horizontal, these forces need to cancel each other out.
The frictional force can be calculated using the formula: f = μk * N
Let's assume the length of the bar is L. At the displaced position, the left side of the bar is displaced by x, and the right side is displaced by (L - x).
Using the normal forces calculated in part (a), the frictional force on the left side is:
f = μk * Nl = μk * mg
The frictional force on the right side is:
f = μk * Nr = -μk * mg
Since the net force must be zero, we have:
-μk * mg - μk * mg = 0
Simplifying this equation gives us the differential equation governing the horizontal displacement:
-2μk * mg = 0
(c) To find the subsequent location of the center of the bar after being released from rest at x = x_0 at t = 0, we need to solve the differential equation obtained in part (b).
The differential equation is:
-2μk * mg = 0
To solve this differential equation, we need to know the initial conditions. The initial conditions are given as x = x_0 at t = 0.
Assuming x = x(t), we can integrate the differential equation with respect to time (t) to find the function x(t).
Integrating the left side of the equation gives us:
∫ -2μk * mg dt = ∫ 0 dt
-2μk * mg * t = C
Where C is the constant of integration.
Applying the initial condition, x = x_0 at t = 0, we can solve for C:
-2μk * mg * 0 = C
C = 0
Therefore, the equation for the horizontal displacement of the bar is:
x = 0
This means that the subsequent location of the center of the bar will remain at the initial position, x = x_0.