Question

A ring of mass (2pi)kg and of radius 0.25m is making 300rp m about an axis through its perpendicular to its plane. The tension in newton developed in ring is approximately

Answer 1

The tension developed in ring is:

• Given, mass, m = 2$\pi$ kg

                    radius, r = 0.25 m

                   number of revolutions, n = 300 rpm

• See the attachment for the resolution of tension. Horizontal forces and vertical forces are balances.

                $\frac{mv^2}{r} =$ 2T sin θ = (dm)rω²

• dm = ( m/l) * θr  

• ω = (2 $\pi$ N)/60
• As θ is small, sin θ = θ

• Solving the above equations, we get : $T = \frac{2 * w^2 * r}{2 \pi }$

• Putting the values, we get : T = 250 N

Answer 2

Answer:

Tension   T=  l mr  2  w  2

  where  m,r,l,w are the mass , radius, circumference and angular speed of the ring respectively.

Given:  m=2πkgr=0.25m

Also  ν=300rpm=  60 300  =5s  −1

 Thus   w=2π(5)=10πrad/s

T=  2π(.25)

2π(.25)  

2 (10π)  2 N

T=246.5N