Answer to Question #96939 in Classical Mechanics for Shalini Bujarbaruah

Question

A disc of radius 12 m is rotating in the clockwise direction with an angular speed of 10 rad/sec. A small, smooth ball bearing capable of sliding along the rectangular slot moves with a speed of 10m/s with respect to an observer at O. The mass of the ball is 100gm. The perpendicular distance of the slot from the centre of disc is 8m. Moreover OP=10 m at an instant t. What is the relative velocity of the ball bearing for an observer on the ground at that instant? What is the centrifugal force on the ball? What is the coriolis force on the ball?

Answer 1

$\omega=10\ rad\ sec^{-1}\\m=0.1\ kg\\v_{bo}=10\ m\ sec^{-1}$

distance of observer from centre at that instant = 10 - 8 = 2m

relative velocity of ball wrt the observer on ground = velocity of ball wrt observer at O + velocity of observer wrt ground

So,

$\overset{\to}v_{bg} =\overset{\to}v_{bo}+\overset{\to}v_{og}$

$v_{bo}=10\ m\ sec^{-1}\\$ anticlockwise

$v_{og}=angular\ velocity\times 2=20 \\$ clockwise

so,

Relative velocity of ball wrt ground=

$v_{bg}=20-10=10\ m\ sec^{-1}\ clockwise$

Centrifugal force on the ball =

$$\frac{m_{ball}v^2_{bo}}r=\frac{0.1\times10^2}{8}=1.25\ N$$

Coriolis force =

$$=2m(v_{bo}\times \omega)\\=2\times0.1(10\times10sin 90\degree)\\=20\ N$$

$\because \ (angle\ between\ v\ and\ \omega\ is\ 90\degree)$