A bar of mass m 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 D. There is a coefficient of kinetic friction μk 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 g.
(a) Find the normal forces NL and NR exerted by the left and right rollers on the bar when the center of the bar is displaced a distance x from the position midway between the rollers. Express your answers in terms of m, x, d and g (enter m for m, x for x, d for d and g for g).
NL=
NR=
(b) Find the differential equation governing the horizontal displacement of the bar x(t). Express your answer in terms of x, d, μk and g (enter x for x, d for d, mu_k for μk and g for g).
d2x/dt2=
(c) The bar is released from rest at x=x0 at t=0. Find the subsequent location of the center of the bar, x(t). Express your answer in terms of x0, d, μk, t and g (enter x_0 for x0, d for d, mu_k for μk, t for t and g for g).
x(t)=
(a)NL = (m*g*(1-(2*x)/d))/2
NR = (m*g*(1+(2*x)/d))/2
(b)(-2*mu_k*g*x)/d
(c)x_0*cos(sqrt((2*mu_k*g)/d)*t)
thanks
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To solve this problem, let's break it down step by step.
(a) To find the normal forces NL and NR exerted by the left and right rollers on the bar when the center of the bar is displaced a distance x, we need to consider the forces acting on the bar.
First, let's understand the forces acting on the bar. We have the weight of the bar, which is mg and acts vertically downward. Since the bar remains horizontal, the normal forces NL and NR act in the vertical direction to balance the weight.
When the bar is displaced a distance x from the position midway between the rollers, the normal force on the left side NL is greater than the normal force on the right side NR. This is because the left side of the bar is displaced downward and experiences an additional downward force due to the weight of the displaced portion.
The normal forces NL and NR can be calculated using the equilibrium condition in the vertical direction.
NL + NR = mg (Equation 1)
To relate the displacement x to the normal forces, we can consider the geometry of the problem. The distance between the center of the bar and the left roller is d/2, and the distance between the center and the right roller is also d/2.
Using similar triangles, we can write the following relationship:
(x + d/2) / (d/2) = NL / (mg) (Equation 2)
Simplifying Equation 2, we get:
NL = (x + d/2) / d * mg
Similarly, we can write the relationship for NR:
(x - d/2) / (d/2) = NR / (mg)
Simplifying Equation 3, we get:
NR = (x - d/2) / d * mg
Therefore, the normal forces NL and NR exerted by the left and right rollers on the bar when the center of the bar is displaced a distance x are:
NL = (x + d/2) / d * mg
NR = (x - d/2) / d * mg
(b) To find the differential equation governing the horizontal displacement of the bar x(t), we need to consider the forces acting in the horizontal direction.
The only force in the horizontal direction is the friction force between each roller and the bar. The friction force can be calculated using the coefficient of kinetic friction μk and the normal force. The kinetic friction force is given by:
F_friction = μk * N (Equation 4)
Since there is no acceleration in the vertical direction, the forces NL, NR, and the weight mg balance each other:
NL - NR = 0 (Equation 5)
Using Equation 4 and Equation 5, we can write:
μk * NL - μk * NR = 0
Substituting the expressions for NL and NR from part (a) into the above equation, we get:
μk * [(x + d/2) / d * mg] - μk * [(x - d/2) / d * mg] = 0
Simplifying, we have:
(x + d/2) - (x - d/2) = 0
This simplifies to:
d = 0
Since this equation is not dependent on x, this means that the horizontal displacement x(t) is independent of time and remains constant.
Therefore, the differential equation governing the horizontal displacement of the bar x(t) is:
d^2x/dt^2 = 0
(c) Since the bar remains at a constant horizontal displacement x, its subsequent location x(t) will be equal to the initial displacement x0.
x(t) = x0