Question

One end of a massless spring of spring constant $k=200\text{N/m}$ and natural length $0.5\text{m}$ is fixed at the centre of a frictionless horizontal table. The other end is connected to a load of mass $1\text{kg}$ lying on the table at rest. If the spring remains horizontal and mass is made to rotate at an angular speed of $2\text{rad/s}$, find the elongation in the spring.

• A. $2\text{cm}$
• B. $1\text{cm}$
• C. $4\text{cm}$
• D. $0.1\text{cm}$

Answer 1

The correct option is B $1\text{cm}$

When the mass $m$ is moving on a circular path around the vertical axis, the spring undergoes elongation to provide the necessary centripetal force for the motion.

Body is in equiibrium in vertical direction ($N=mg$). So we only have to consider the motion on the horizontal plane.

Let the elongation produced in the spring be $x\text{metres}$
i.e Spring force ($F$) = $kx$
Radius of circular path $r=$ natural length of spring $=0.5\text{m}$

Since centripetal force is provided by the spring force,
$kx=mr{\omega }^2$
$\Rightarrow x=\frac{mr{\omega }^2}{k}=\frac{1\times 0.5\times (2{)}^2}{200}=\frac{2}{200}\text{m}=1\text{cm}$

NOTE: Elongation in the spring is very small with respect to the natural length of the spring i.e $x<<r$. Hence our assumption of radius of circular path $=r$, holds good.