First slide

Dynamics of uniform and non uniform circular motion

Question

A spring mass system ( mass m , spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc together with spring mass system, rotates about it’s axis with an angular velocity $ω, (k > > m ω^{2})$ the relative change in the length of the spring is best given by the option

Moderate

A

$\frac{2 m ω^{2}}{k}$
B

$\frac{m ω^{2}}{k}$
C

$\sqrt{\frac{2}{3}} (\frac{m ω^{2}}{k})$
D

$\frac{m ω^{2}}{3 k}$

Solution

Spring force=centrifugal force
$k x = m (l + x) ω^{2}$
$\Rightarrow k x - m x ω^{2} = m l ω^{2} \Rightarrow x = \frac{m l ω^{2}}{k - m ω^{2}} \Rightarrow x = \frac{m l ω^{2}}{k} [k > > m ω^{2}]$
Relative change is $\frac{x}{l} = \frac{m ω^{2}}{k}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App