A ring of mass 2 π kg and of the radius 0.25 m is making 300 rpm about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

Question

A ring of mass 2 π kg and of the radius 0.25 m is making 300 rpm about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As 2 textt sin (θ /2) = textdm ω 2 textr (for small angle sin (θ /2) arrow (θ /2) ) but textdm = ( textm/ℓ ) θ textr As ℓ = 2 π textr ∴ textT = ( textm ω 2 textr/2 π ) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-t4vid6ox8jpucpch.jpg" /> Put textm = 2 π kg text, ω = 1 0 π rads- 1 and r=0.25m ∴ textT = 2 5 0 textN

Q. A ring of mass $2 πk g$ and of the radius $0.25 m$ is making $300 r p m$ about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

As $2 t s in \frac{θ}{2} = dm ω^{2} r$ (for small angle sin $\frac{θ}{2} \to \frac{θ}{2}$ )
but $dm = \frac{m}{ℓ} θ r$
As $ℓ = 2 π r$
$∴ T = \frac{m ω ^{2} r}{2 π}$

Put $m = 2 πk g, ω = 10 π$ $r a d s^{- 1}$ and $r = 0.25 m$
$∴ T = 250 N$

Q. A ring of mass 2πkg and of the radius 0.25m is making 300rpm about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

As 2tsin2θ​=dmω2r (for small angle sin 2θ​→2θ​ ) but dm=ℓm​θr As ℓ=2πr ∴T=2πmω2r​ Put m=2πkg, ω=10π rads−1 and r=0.25m ∴T=250N

Q. A ring of mass $2 πk g$ and of the radius $0.25 m$ is making $300 r p m$ about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

As $2 t s in \frac{θ}{2} = dm ω^{2} r$ (for small angle sin $\frac{θ}{2} \to \frac{θ}{2}$ )
but $dm = \frac{m}{ℓ} θ r$
As $ℓ = 2 π r$
$∴ T = \frac{m ω ^{2} r}{2 π}$

Put $m = 2 πk g, ω = 10 π$ $r a d s^{- 1}$ and $r = 0.25 m$
$∴ T = 250 N$