Question

A table with smooth horizontal surface is turning at an angular speed $\omega$ about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle as its distance from the centre becomes L.

• A. v = L
• B. v = a
• C. v =
• D. v =

Answer 1

The correct option is C

v =

The situation is shown in figure.

Let us work from the frame of reference of the table.

Let us take the origin at the centre of rotation 0 and the X-axis along the groove. Suppose at time t the

particle in the groove is at a distance x from the origin and is moving along the X-axis with a speed v. The

forces acting on the particle (including the pseudo forces that we must assume because we have taken

our frame on the table which is rotating and is nonintertial) are

(a) weight mg vertically downward,

(b) normal contact force $N_1$ vertically upward by the bottom surface of the groove,

(c) centrifugal force m${\omega }^2$x along the X-axis

As the particle can only move in the groove, its acceleration is along the X-axis. The only force along the

X-axis is the centrifugal force m${\omega }^2$x. All the other forces are perpendicular to the X-axis and have

no components along the X-axis.
$F=m{\omega }^{2}x$
$a={\omega }^{2}x$
$v\frac{dv}{dx}={\omega }^{2}x$
$\Rightarrow vdv={\omega }^{2}xdx$
Integrating on both side
$\Rightarrow {\int }_0^{v}vdv={\omega }^2{\int }_a^{L}xdx\Rightarrow v=\omega \sqrt{L^2-a^2}$