A body of mass m= 20 g is attached to an elastic spring of length L=50 cm and spring constant k=2 `Nm^(-1)`. The syste is revolved in a horizontal plan with a frequency v=30 rev/min. Find the radius of the circular motion and the tension in the spring .

Options

A. 0.25 m, 0.1 N
B. 0.5 m, 0.52 N
C. 0.55 m 0.1 N
D. 0.9 m, 0.2 N

Answer

C

Explanation

Angular velocity `omega=2pir=2pixx(30)/(60)=pirad//s`
For an elastic spring force `F=k_(x)` where x is the extension.
Radius of circular motion r=L+x
Centripetal force `=mromega^(2)=F`
`impliesm(L+x)omega^(2)=Kx`
`impliesx=(mLomega^(2))/(k-momega^(2))=(0.02xx0.5xx(3.14)^(2))/(2-0.02xx(3.14)^(2))`
`~~0.05m`
Radius of the circular motion (r)
`=L+x=0.5+0.05`
=0.55 M
Tension in the spring
`T=kx=2xx0.05~~0.1N`