A bead of mass m can slide without friction on a fixed circular horizontal ring of radius 3R having centre at the point C. The bead is attached to one of the ends of spring of spring constant k. Natural length of spring is R and the other end of the spring is fixed at point O as shown in figure. Bead is released from position A, what will be kinetic energy of the bead when it reaches at point B ?
- $12 \, k R^2$
- $\frac{25}{2} k R^2$
- $\frac{9}{2} k R^2$
- $ 8 k R^2$
The Correct Option is D
Solution and Explanation
mass of bead $=m$
radius of circular horizontal ring $=3 R$
natural length of spring $=R$
According to conservation law of energy,
$ K E_{i}+P E_{i}=K E_{f}+P E_{f} $
$ \Rightarrow 0+\frac{1}{2} k[O A-R]^{2}=K E_{f}+\frac{1}{2} k[O B-R]^{2} $
$ \Rightarrow \, \frac{1}{2} k[5 R-R]^{2}=K E_{f}+\frac{1}{2} k[R-R]^{2}$
$ \Rightarrow \, K E_{f}=8 k R^{2} $
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Concepts Used:
Kinetic energy
Kinetic energy of an object is the measure of the work it does as a result of its motion. Kinetic energy is the type of energy that an object or particle has as a result of its movement. When an object is subjected to a net force, it accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving object or particle defined by both its mass and its velocity. Any combination of motions is possible, including translation (moving along a route from one spot to another), rotation around an axis, vibration, and any combination of motions.
