Question

A particle of mass 10g moves along a circle pf radius 6.4 cm with a constant tangential accelaration. What is the magnitude of this accelaration if the kinetic energy of the particle becomes equal to 8×10^-4 by tye end of the second revolution after the beginning of the motion?

Answer 1

Given
mass of particle m = 0.01 kg
radius of circle along which particle is moving , r= 6.4 cm

kinetic energy of the particle K. E. = 8×10<sup>-4 J

1/2 mv^2 =</sup> 8×10<sup>-4 J

v^2=</sup> (16×10^-4 ) /0.01

v^2= 16*10^-2 -------- (i)
Given that K. E. Of particle is equal to 8×10^-4
by tye end of the second revolution after the beginning of the motion of particle.
it means initial velocity (u ) 0 m/s at this moment.

now using newton's 3rd equation of motion ,
v^2 = u^2 + 2as
v^2= 2as = 2 a (4 pi r)
a= v^2/8 pi r
a = (16*10^-2) / (8*3.14*6.4*10^-2)
a = 0.1 m/s^r

ans= 0.1 m/s^2