A 13 collar C slides on a horizontal rod between springs A

A 13 # collar C slides on a horizontal rod between springs A and B. If the collar is pushed to the left until spring A is compressed 1.5 in and released, determine the distance through which the collar will travel assuming No friction between the collar and the rod A coefficient of friction mu_k = 0.3 A coefficient of friction mu_k = 0.08

Solution

As the system has no external force acting on it, we will use the equation for the conservation of energy to determine the distance travelled by the collar in the three cases mentioned. It needs to be noted that for a spring with spring constant of k and compression of x the energy stored is given as 0.5kx^2 while for a frictional force of f acting over a distance of x, the work done on the body is fx.

When the spring A is compressed by 1.5in, the energy stored in the spring A would be:

E = 0.5 x 18 x 2.25 = 20.25 Units

Part a.) Now, on being released the collar will travel towards the other end of the rod and hit the spring B and at maximum compression of spring B the energy stored in it must be equal to the initial energy of the system.

That is 0.5 *12 * x^2 = 20.25

or, X^2 = 3.375

or, X = 1.83712 inches.

That is the spring B is compressed by a distance of 1.83712 inches. Therefore the distance travelled by the collar would be:10 + 1.5 + 1.83712 = 1.83712 inches [Since the collar itself is of 6 inches, we wil have to take 16 - 6 for the distance travelled along the empty section in between]

Part b.) Here, we will first determine the energy loss by the time collar reaches the spring B and then check for further displacement.

E(initial) = 20.25

Work done by friction while it travels for 11.5 inches to hit B is 13 x 0.3 x 11.5 = 44.85

That would mean that the friction will exhaust its energy well before it reaches the spring B

Hence 13*0.3*x = 20.25

or, X = 5.19231 Inches is the required distance travelled.

Part c.) As the friction here lower than the previous case, the collar will manage to hit the spring B.

We will assume that the spring B is compressed by a distance of x.

That would mean that the initial energy is exhausted in compressing the spring as well as by the work the friction would do while it travels along the rod, hence we have:

20.25 = 13 (0.08) x (11.5 + x) + 0.5*12*x^2

or, 8.24 = 1.04x + 6x^2

or, x = [-1.04 +/- sqrt(1.04*1.04 + 24(8.24))] / 12 = [-1.04 +/- 14.1011] / 12 = 1.088 inches

That is the spring B is compressed by 1.088 inches.

Hence the total distance travelled would be: 11.5 + 1.088 = 12.5884 inches