Question

Where on the rope do you expect the tension will be greatest? Where do you expect it will be the least? A uniform rope with length L and mass m is held at one end and whirled in a horizontal circle with angular velocity ω. You can ignore the force of gravity on the rope. (a) At a point on the rope a distance r from the end that is held, what is the tension F? (b) What is the speed of transverse waves at this point? (c) Find the time required for a transverse wave to travel from one end of the rope to the other.

Answer 1

a)
`F=(\omega^2 m)/(2L)\left({L^2}-{r^2}\right)`

b)
`V=\omega \sqrt{(L^2-r^2)/(2)}`

c)
`t=(\pi)/(\omega \sqrt2)`

Step-by-step explanation:

Given that

Length =L

Mass = m

Force on elemental part

dF= dm ω² r

dm = m/L dr

dF= ω² r m/L dr

By integrating from r to L

`F=(m)/(L) \int_(r)^(L)\omega^2 r \ dr`

`F=(\omega^2 m)/(L)\left((L^2)/(2)-(r^2)/(2)\right)`

`F=(\omega^2 m)/(2L)\left({L^2}-{r^2}\right)`

Velocity V

`V=\sqrt{(F)/((m)/(L))}`

`V=\sqrt{\frac{(\omega^2 m)/(2L)\left({L^2}-{r^2}\right)}{(m)/(L)}}`

`V=\omega \sqrt{(L^2-r^2)/(2)}`

`(dr)/(dt)=V=\omega \sqrt{(L^2-r^2)/(2)}`

`\int_(0)^(L)(dr)/(√(L^2-r^2))=\int_(0)^(t) (\omega )/(\sqrt2)dt`

`sin^(-1)1=\ (\omega )/(\sqrt2)t`

`(\pi)/(2)=\ (\omega )/(\sqrt2)t`

`t=(\pi)/(\omega \sqrt2)`