Question

A light spiral spring has an unstretched length 1 and a
spring constant k. A small object of mass m is attached
a vertical axis that passes through a small light ring
attached to the other end of the spring as shown in figure

If the object travels along a circular path of radius R
with constant angular speed w, keeping the spring on a
Horizontal plane, then prove that w is equal to
$\sqrt{k \m}$

​

Answer 1

Given :

Spring constant of the spring = k

Mass of small object = m

Radius of circular path = R

Angular velocity of object = ω

To Prove :

We have to prove that angular velocity (ω) of object is √(k/m)

Solution :

Here the concepts of centripetal force and spring force are used.

• We know that centripetal force always acts on the body towards the centre of circle in any kinda circular motion.

❖ Centripetal force is not a new kind of force. The material force such as tension, gravitational force, electrical force, friction, spring force, etc., may act as the centripetal force in any circular motion.

• Centripetal force is the name give to any force that provides radial inward acceleration to a body in circular motion.

$\sf:\implies\: Centripetal\:force=Spring\:force$

$\sf:\implies\:\dfrac{mv^2}{R}=k\cdot R$

• We know that, v = R ω

$\sf:\implies\:{m(R\omega)^2}=k\cdot R^2$

$\sf:\implies\:mR^2\omega^2=k\cdot R^2$

$\sf:\implies\:m\omega^2=k$

$\sf:\implies\:\omega^2=\dfrac{k}{m}$

$:\implies\:\underline{\boxed{\bf{\orange{\omega=\sqrt{\dfrac{k}{m}}}}}}$