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Question

One end of a massless spring of spring constant k=200 N/m  and natural length 0.5 m is fixed at the centre of a frictionless horizontal table. The other end is connected to a load of mass 1 kg lying on the table at rest. If the spring remains horizontal and mass is made to rotate at an angular speed of 2 rad/s, find the elongation in the spring.
  1. 2 cm
  2. 1 cm
  3. 4 cm
  4. 0.1 cm


Solution

The correct option is B 1 cm
When the mass m is moving on a circular path around the vertical axis, the spring undergoes elongation to provide the necessary centripetal force for the motion.

Body is in equiibrium in vertical direction (N=mg). So we only have to consider the motion on the horizontal plane.

Let the elongation produced in the spring be x metres
i.e Spring force (F) = kx 
Radius of circular path r= natural length of spring =0.5 m

Since centripetal force is provided by the spring force,
kx=mrω2
x=mrω2k=1×0.5×(2)2200=2200 m=1 cm

NOTE: Elongation in the spring is very small with respect to the natural length of the spring i.e x<<r. Hence our assumption of radius of circular path =r, holds good.

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