Question

One end of a massless spring of spring constant $100N/m$ is fixed at the centre of a frictionless horizontal table and the other end is connected to a body of mass $7kg$ lying on the table at rest. The spring has a natural length of $2m$. If the spring remains horizontal and mass is made to rotate at an angular speed of $1rad/s$, then find the elongation in the spring, assuming the elongation to be small compared to the natural length of the spring.

• A. $14cm$
• B. $3.5cm$
• C. $35cm$
• D. $1.4cm$

Answer 1

The correct option is A $14cm$

When the body is rotating in a horizontal circular path, then the spring undergoes elongation to provide the required centripetal force for the body.

Applying the equation of circular motion along radial direction
$F_{net}=mr{\omega }^2$ .....(1)

Now let us assume the extension in spring as $x$ metres.
$\therefore$ Spring force $F_{spring}=kx$

$\Rightarrow kx=mr{\omega }^2$ ....(2)

Substituting $m=7kg,\omega =1rad/s$
$r=$ natural length of spring $=2m$ in equation (2)

$\therefore$ $x=\frac{mr{\omega }^2}{k}=\frac{7\times 2\times (1{)}^2}{100}=0.14m=14cm$