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A paritcal of mass 10 g moves along a ci...

A paritcal of mass `10 g` moves along a circle of radius `6.4 cm` with a constant tangennitial acceleration. What is the magnitude of this acceleration . What is the magnitude of this acceleration if the kinetic energy of the partical becomes equal to `8 xx 10^(-4) J` by the end of the second revolution after the beginning of the motion?

A

`0.15 m//s^(2)`

B

`0.18m//s^(2)`

C

`0.20 m//s^(2)`

D

`0.10 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Here, `m = 10g = 10^(-2) kg`
`r = 6.4 cm xx 10^(-2) m`
`a_(t) = ?`
`K.E. = (1)/(2)mv^(2) = 8 xx 10^(-4) J`
`v =sqrt((2 xx 8 xx 10^(-4))/(10^(-2))) = 0.4 m//s`
This is the velocity at the end of two revolutions after starting from rest, i.e. `u = 0, s = 2 xx 2 pi r`
From `v^(2) = u^(2) + 2a_(t)s`
`v^(2) = 2a_(t) s` (`:' u = 0)`
`a_(t) = (v^(2))/(2s) = (0.4 xx 0.4)/(2 xx 2 xx 2pi r)`
`a_(t) = (16 xx 10^(-2))/(8 xx 3.14 xx 6.4 xx 10^(-2)) = 0.1 m//s^(2)`
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