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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Motion in Two Dimensions

JIPMER 2015 · Physics · Motion in Two Dimensions

One end of a massless spring of constant 100 ~N / m and natural...

One end of a massless spring of constant 100 N/m100 \mathrm{~N} / \mathrm{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 2rad/s2 \mathrm{rad} / \mathrm{s}, then elongation of spring is

Options

  1. A0.1 m
  2. B10 cm
  3. C1 cm
  4. D0.01 cm

Correct answer

C. 1 cm

Step-by-step solution

For the circular motion acceleration towards the centre is v2r\frac{v^2}{r}. The horizontal force on the particle is due to the spring and klk l, where ll is the elongation and kk is spring constant of spring. kl=mv2r=mω2r=mω2(l0+l)⇒(k−mω2)l=mω2l0⇒l=mω2l0k−mω2\begin{aligned} & k l=\frac{m v^2}{r} & =m \omega^2 r=m \omega^2\left(l_0+l\right) \\ \Rightarrow \quad & \left(k-m \omega^2\right) l & =m \omega^2 l_0 \\ \Rightarrow & l & =\frac{m \omega^2 l_0}{k-m \omega^2}\end{aligned} Putting the values 1=0.5×4×0.5100−0.5×4=198 m=1 cm1=\frac{0.5 \times 4 \times 0.5}{100-0.5 \times 4}=\frac{1}{98} \mathrm{~m}=1 \mathrm{~cm}
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