Question

A thin ring of radius R is made of a material of density rhoand Young's modulus Y. If the ring is rotated about its centre in its own plane with angular velocity omega , find the small increases in its radius.

Answer 1

Answer:

I don't know the answer

Answer 2

Given:-

Radius of ring R

Young's Modulus Y.

To Find:-

Change in radius.

Solution:-

Please find the attached file.

Consider an element PQ of length dl, Let T be the tension & A be the area of cross section of wire.

The component of T , towards the center provides the necessary centripetal force to position PQ.

Therefore Net Force:

F = $Tsin\frac{d\theta }{2}} + Tsin\frac{d\theta }{2}}$

F = $2Tsin\frac{d\theta }{2}}$

dθ very small sinθ≅ θ

Then Net force F = $Tsin\frac{d\theta }{2}}$ ≅ 2T(dθ/2)

F = Tdθ..............(1)

Therefore the second way of centripetal force

F =  = dmrw² = ρ(A)dlRw²

F = ρAdlRw²............(2)

Com-pairing equation 1 and 2

Tdθ = ρAdlRw²           Therefore dl = Rdθ

Tdθ = ρARdθRw²

dθ cancel out both side

Then we get:

Tension Tdθ = AR²w²ρ

Now we have to find the change in Radius ΔR ,

We have strain = $\frac{Change in length}{Original length}$ = Δ2πR/2πR = Δl/l

Strain = $\frac{2\pi \Delta R}{2\pi R} = \Delta\frac{l}{l}$

$\Delta\frac{l}{l} = \Delta\frac{R}{R}$

Now According to Young's Modulus Y = $\frac{Tension }{Cross Section Area} = \frac{\frac{T}{A} }{\frac{\Delta l}{l} }$

Now Using strain value:

Y = $\frac{\frac{T}{A} }{\frac{\Delta R}{R} } = \frac{TR}{A\Delta R}$

Y = $\frac{AR^{2}w^{2}p R}{A\Delta R} = \frac{R^{2}w^{2}pR }{\Delta R}$

Now we get:

ΔR = $\frac{R^{3}w^{2}p }{Y}$.....This is the small increases in its radius