Question

A small particle of mass m is given an initial high velocity in the horizontal plane and winds its cord around the fixed vertical shaft of radius a. Motion occurs essentially in horizontal plane (assume gravity free space). The angular velocity of the cord is ${\omega }_0$ when the distance from the particle to the tangency point is $r_0$.

• A. The angular velocity of the cord after it has turned through an angle is
• B. The angular velocity of the cord after it has turned through an angle is
• C. The KE of the particle will increase
• D. The KE of the particle will remain same

Answer 1

Answer: (A), (D)

The correct options are
A

The angular velocity of the cord after it has turned through an angle is

D

The KE of the particle will remain same

Speed v will remain same, as T is perpendicular to v always. Hence KE remains same.

$v={\omega }_0{r}_0=\omega (r_0-a\theta )$

$\Rightarrow \omega =\frac{{\omega }_0{r}_0}{r_0-a\theta }=\frac{{\omega }_0}{1-\frac{a\theta }{r_0}}$