Question

One end of a massless sprinf of spring constant 100 N/m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on as frictionless horizontla table. The spring remains horizontal. If the mass is made to rotate at an angular velocityof 2 rad/s, find the elongation of the spring.

Answer 1

The particle is moving in a horizontal circle so it si accelerated towards the centre with magnitude `v^2/r`. The horizontal foce on the particle is due toteh spring and equals kl, where l is the elongation and k is the spring constant. Thus
`kl=mv^2/r=momega^2=momega^2(l_0+l)`.
Here omega is the angular velocity `l_0` is theN/Atural length (0.5m) and `l_0+l` is the total length of the spring which is also the radius of the circle along which the particle moves.
`Thus, (k-momega^2)=momega^2l_0`
`or l=(momega^2l_0)/(k-momega^2)`
Putting the values
`l=(0.5xx4xx0.5)/(100-0.5xx4)m~~1/100m=1cm`.