Question

A heavy rope of mass m and length 2L is hanged on a smooth little peg with equal lengths on two sides of the peg. Right part of the rope is pulled a little longer and released. The rope begins to slide under the action of gravity. There is a smooth cover on the peg (so that the rope passes through the narrow channel formed between the peg and the cover) to prevent the rope from whiplashing.

(a) Calculate the speed of the rope as a function of its length (x) on the right side. (b) Differentiate the expression obtained in (a) to find the acceleration of the rope as a function of x. (c) Write the rate of change of momentum of the rope as a function of x. Take downward direction as positive (d) Find the force applied by the rope on the peg as a function of x. (e) For what value of x, the force found in (d) becomes zero? What will happen if there is no cover around the peg?

Answer 1

(a) $v(x) = \sqrt{\frac{g}{L}}(x-L)$ for $x \ge L$ (b) $a(x) = \frac{g}{L}(x-L)$ (c) $\frac{dp}{dt}(x) = m\frac{g}{L}(x-L)$ (d) $F_{peg}(x) = \frac{mgx(2L-x)}{L^2}$ (e) The force becomes zero for x=0 and x=2L. If there is no cover around the peg, the rope might lose contact with the peg (whiplash) if the required normal force is not provided, leading to a different motion than described.

Answer 2

The problem involves a rope sliding over a peg, and we need to find expressions for its speed, acceleration, momentum change, and the force on the peg, along with a discussion of what happens without a cover on the peg.

Detailed Solution:

(a) Speed as a function of x:

Using conservation of energy, we find the potential energy difference as the rope slides. The initial potential energy is $U_i = -mg\frac{L}{2}$. At a later time, the potential energy $U(x)$ is given by:

$$U(x) = -\frac{mg}{2L}[x^2 - 2Lx + 2L^2]$$

The kinetic energy is $K(x) = \frac{1}{2}mv^2$. Applying conservation of energy:

$$\frac{1}{2}mv^2 - \frac{mg}{2L}[x^2 - 2Lx + 2L^2] = -mg\frac{L}{2}$$

Solving for $v$:

$$v(x) = \sqrt{\frac{g}{L}}(x-L)$$

This is valid for $x \ge L$.

(b) Acceleration as a function of x:

The acceleration $a$ is the derivative of velocity with respect to time. Using the chain rule:

$$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$$

Given $v(x) = \sqrt{\frac{g}{L}}(x-L)$, then $\frac{dv}{dx} = \sqrt{\frac{g}{L}}$.

Thus, the acceleration is:

$$a(x) = \frac{g}{L}(x-L)$$

(c) Rate of change of momentum:

The rate of change of momentum is $\frac{dp}{dt} = ma$. Using the acceleration from part (b):

$$\frac{dp}{dt}(x) = m\frac{g}{L}(x-L)$$

(d) Force applied by the rope on the peg:

The tension in the rope is $T(x) = \frac{mgx(2L-x)}{2L^2}$. The force applied by the rope on the peg is twice the tension:

$$F_{peg}(x) = 2T(x) = \frac{mgx(2L-x)}{L^2}$$

(e) Value of x for zero force and effect of no cover:

The force is zero when $x(2L-x) = 0$, which gives $x = 0$ or $x = 2L$.

If there is no cover, the rope may lose contact with the peg (whiplash) if the required normal force is not provided, leading to a different motion. The cover ensures the rope stays in contact, providing additional constraint force if needed.