Question

A chain of 125 links is 1.25 m long and has a mass of 2 kg with the ends fastened together it is set rotating at $3000\frac{rev}{min}.$ Find the centripetal force on each link-

• A. $3.14N$
• B. $314N$
• C. $\frac{1}{314}N$
• D. $\frac{1}{3.14}N$

Answer 1

The correct option is B $314N$

Given,

Angular velocity,

$\omega =3000rmp$

$\omega =\frac{2\pi }{60}\times 3000=100\pi rad/s$

No. of link in chain is $125$ of mass $2kg$

Mass of each link, $m=\frac{2}{125}=0.016kg$

Length of cirular path, $l=2\pi r=1.25m$

$r=\frac{l}{2\pi }=\frac{1.25}{2\pi }=0.1999m$

The centripetal force,

$f_r=m{\omega }^{2}r$

$f_r=0.016\times (100\pi {)}^2\times 0.1999$

$f_r=314N$

The correct option is B.