A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l . The system is rotated about the other end of the spring with an angular velocity ω , in gravity-free space. Then increase in the length of the spring will be:

Question

A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l . The system is rotated about the other end of the spring with an angular velocity ω , in gravity-free space. Then increase in the length of the spring will be: (A) (m ω 2 l/k) (B) (m ω 2 l/k - m ω 2) (C) (m ω 2 l/k + m ω 2) (D) None of these

Tardigrade Admin · Accepted Answer

Let x be increase in length of the spring. The particle would move in a circular path of radius (.l+x.) . Centripetal force = force due to the spring m(l + x)(ω )2=kx ∴ x=(m ω 2 l/k - m ω 2)

Q. A particle of mass $m$ is fixed to one end of a light spring of force constant $k$ and unstretched length $l$ . The system is rotated about the other end of the spring with an angular velocity $ω$ , in gravity-free space. Then increase in the length of the spring will be:

$\frac{m ω ^{2} l}{k}$

$\frac{m ω ^{2} l}{k - m ω ^{2}}$

$\frac{m ω ^{2} l}{k + m ω ^{2}}$

None of these

Let $x$ be increase in length of the spring. The particle would move in a circular path of radius $(l + x)$ . Centripetal force = force due to the spring
$m (l + x) (ω)^{2} = k x$
$∴ x = \frac{m ω ^{2} l}{k - m ω ^{2}}$

Q. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l . The system is rotated about the other end of the spring with an angular velocity ω , in gravity-free space. Then increase in the length of the spring will be:

kmω2l​

k−mω2mω2l​

k+mω2mω2l​