One end of a massless spring of constant 100 N /m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at angular velocity of 2 rad/s, then elongation of spring is

0.1 m

10 cm

1 cm

0.01 cm

Solution

1 cm

For the circular motion acceleration towards the centre is $\frac{v^{2}}{r}$ .

The horizontal force on the particle is I due to the spring and kl,

where, l is the elongation

k is spring constant of spring.

kl = $\frac{{mv}^{2}}{r}$

= mω² r

kl = mω² ( l_o + l )

⇒ kl - mω² ( l_o + l ) = 0

⇒ ( k - mω²) l = mω² l_o

⇒ l = $\frac{{mω}^{2} l_{o}}{k - {mω}^{2}}$

Putting tle values l = $\frac{0.5 \times 4 \times 0.5}{100 - 0.5 \times 4}$

= $\frac{1}{98}$ m

= 0.010 m

l = 1 cm

Hence elongation(l) in spring is 1 cm.

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