A bead of mass m can freely slide
down the fixed inclined rod without
friction. It is connected to a point P on
the horizontal surface with a light
spring of spring constant k. The bead is
initially released from rest and the
spring is initially unstressed and
vertical. The bead just stops at the
bottom of the inclined rod. Find the
angle which the inclined rod makes
with the horizontal.
m
MITTITI
Р
Answers
The kinetic energy of the bead will be K.E f=8 k R^2
Explanation:
Correct statement:
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?
Given data:
- Mass of bead = m
- Radius of circular horizontal ring = 3R
- Natural length of spring = R
According to conservation law of energy,
K.E i +P.E i =K.E f +P.E f
0 + 1/2 k[O.A−R]^2 = K.E f + 1/2 k[OB − R]^2
1 − k [5R − R] ^2 = =K.E f + 1/2 k[R−R]^2
K.E f=8 k R^2
Thus the kinetic energy of the bead will be K.E f=8 k R^2
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