Question

A table with smooth horizontal surface is rotating at a speed ω 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 with respect to the table as its distance from the centre becomes l.

• A. v = ωl
• B. v = ω(l - a)
• C. v = $\frac{\omega(l + a)}{2}$
• D. v = ω$\sqrt{l^{2} - a^{2}}$

Answer 1

Answer: (D)

v = ω$\sqrt{l^{2} - a^{2}}$

Answer 2

$\frac{F}{m} = \frac{m\omega^{2}x}{m}or\frac{dv}{dt} = \omega^{2}x$

$\frac{dv}{dx}.\frac{dx}{dt} = \omega^{2}x$

vdv = ω²x dx

$\int_{0}^{v}{vdv} = \int_{a}^{L}{\omega^{2}xdx}$

v = ω$\sqrt{l^{2} - a^{2}}$