Question

A uniform rope of length L and mass M is held at one end and whirled in a horizontal circle with angular velocity 'w'.Find the time required for a transverse wave to travel from one end of the rope to the other

Answer 1

Answer on Question 57878, Physics, Mechanics, Relativity

Question:

A uniform rope of length $L$ and mass $M$ is held at one end and whirled in a horizontal circle with angular velocity $\omega$. You can ignore the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.

Solution:

Let's consider a small segment of the rope between $r$ and $r + \Delta r$. The segment has a length of $\Delta r$ and a mass of $m = M\Delta r / L$. Let's treat this segment as a particle and consider the forces and the Newton's law on this segment. The inward force of tension from the rope to the segment is $T(r)$, and the outward force to the segment is $T(r + \Delta r)$. The acceleration of the segment is $r\omega^2$ and it points toward the pivot. Let's assume that the positive $r$ direction is outward, then the acceleration is negative, and applying the Second Newton's law we get:

$$T(r + \Delta r) - T(r) = -mr\omega^2,$$ $$T(r + \Delta r) - T(r) = -\frac{M}{L}\Delta r r\omega^2,$$ $$\frac{T(r + \Delta r) - T(r)}{\Delta r} = -\frac{M}{L}r\omega^2.$$

Let $\Delta r$ approach zero, then, using the definition of derivative we get:

$$\frac{dT}{dr} = -\frac{M\omega^2}{L}r,$$ $$dT = -\frac{M\omega^2}{L}rdr,$$

After integration we get:

$$\int_{T_0}^{T(r)} dT = -\int_0^r \frac{M\omega^2}{L}rdr,$$ $$T = -\frac{M\omega^2}{L}\frac{r^2}{2} + C,$$

here, $C$ is an integration constant and can be determined by considering the constraint that the force of tension $T$ is zero at the tip of the rope, so $T = 0$ at $r = L$ and we get:

$$C = \frac {M \omega^ {2}}{2} L,$$ $$T = - \frac {M \omega^ {2}}{L} \frac {r ^ {2}}{2} + \frac {M \omega^ {2}}{2} L = \frac {M \omega^ {2}}{2 L} (L ^ {2} - r ^ {2}),$$ $$T = \frac {M \omega^ {2}}{2 L} (L ^ {2} - r ^ {2}).$$

Then, we can find the velocity of the transverse wave in the rope from the formula:

$$v = \sqrt {\frac {T}{\mu}},$$

here, $T$ is the force of tension in the rope, $\mu = M / L$ is the linear mass density of the rope.

Substituting $\mu$ into the previous formula we can find the velocity of the transverse wave in the rope:

$$v = \sqrt {\frac {T}{\mu}} = \sqrt {\frac {\frac {M \omega^ {2}}{2 L} (L ^ {2} - r ^ {2})}{\frac {M}{L}}} = \sqrt {\frac {\omega^ {2}}{2} (L ^ {2} - r ^ {2})},$$ $$\frac {d r}{d t} = \sqrt {\frac {\omega^ {2}}{2} (L ^ {2} - r ^ {2})} = \frac {\omega}{\sqrt {2}} \sqrt {(L ^ {2} - r ^ {2})},$$ $$\frac {d r}{\sqrt {(L ^ {2} - r ^ {2})}} = \frac {\omega}{\sqrt {2}} d t,$$ $$\int_ {0} ^ {L} \frac {d r}{\sqrt {(L ^ {2} - r ^ {2})}} = \frac {\omega}{\sqrt {2}} \int_ {0} ^ {t} d t,$$ $$\arcsin 1 = \frac {\omega}{\sqrt {2}} t,$$ $$\frac {\pi}{2} = \frac {\omega}{\sqrt {2}} t,$$ $$t = \frac {\pi}{\sqrt {2} \cdot \omega}.$$

Answer:

$$t = \frac {\pi}{\sqrt {2} \cdot \omega}$$

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