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Question: A particle of mass 10 grams moves along a circle of radius 6.4 cm with a constant tangential acceler...

A particle of mass 10 grams moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes to 8×10−4J8 \times {10^{ - 4}}J by the end of the second revolution after the beginning of the motion?
A. 0.1 m/s20.1{\text{ m/}}{{\text{s}}^2}
B. 0.15 m/s20.15{\text{ m/}}{{\text{s}}^2}
C. 0.18 m/s20.18{\text{ m/}}{{\text{s}}^2}
D. 0.2 m/s20.2{\text{ m/}}{{\text{s}}^2}

Explanation

Solution

In this question first we need to find the total distance travelled by particle after two revolutions in a circular path, by finding the circumference of the circular path and then multiplying it by two since the particle made two revolutions.

Complete step by step answer: Mass of the particle m=10g=10×10−3kgm = 10g = 10 \times {10^{ - 3}}kg
The radius of the circle r=6.4cm=6.4×10−2mr = 6.4cm = 6.4 \times {10^{ - 2}}m
Kinetic energy =8×10−4J8 \times {10^{ - 4}}J
Where kinetic energy is given by the formulaKE=12mv2−−(i)KE = \dfrac{1}{2}m{v^2} - - (i)
Hence we can write

12mv2=8×10−4J 12(10×10−3)v2=8×10−4J  \dfrac{1}{2}m{v^2} = 8 \times {10^{ - 4}}J \\\ \dfrac{1}{2}\left( {10 \times {{10}^{ - 3}}} \right){v^2} = 8 \times {10^{ - 4}}J \\\

By further solving we get

v2=2×8×10−410×10−3 v2=16×10−2 v=4×10−1 v=0.4 m/s−−(ii)  {v^2} = \dfrac{{2 \times 8 \times {{10}^{ - 4}}}}{{10 \times {{10}^{ - 3}}}} \\\ {v^2} = 16 \times {10^{ - 2}} \\\ v = 4 \times {10^{ - 1}} \\\ v = 0.4{\text{ m/s}} - - (ii) \\\

Now find the total distance covered by the particle after two revolutions by using
S=2(2πr)−−(iii)S = 2\left( {2\pi r} \right) - - (iii)
By solving we get

S=4πr =4×3.14×6.4×10−2 =0.8m  S = 4\pi r \\\ = 4 \times 3.14 \times 6.4 \times {10^{ - 2}} \\\ = 0.8m \\\

Now use the third equation of newton’s law of motion which is given as
v2=u2+2aS−−(iii){v^2} = {u^2} + 2aS - - (iii)
Now since the particle started from the rest, hence its initial velocity will be
u=0u = 0
Now substitute the value of distance S and the velocity v in equation (iii) to find the acceleration; hence we get

v2=u2+2aS−−(iii) (0.4)2=0+2a(0.8) 0.16=2a(0.8) a=0.1 m/s2  {v^2} = {u^2} + 2aS - - (iii) \\\ {\left( {0.4} \right)^2} = 0 + 2a(0.8) \\\ 0.16 = 2a(0.8) \\\ a = 0.1{\text{ m/}}{{\text{s}}^2} \\\

Therefore, the magnitude of this acceleration by the end of the second revolution =0.1 m/s2 = 0.1{\text{ m/}}{{\text{s}}^2}
Option A is correct

Note: The newton’s third equation of motion establishes a relation between the initial velocity, final velocity, acceleration and the displacement of the particle which is given as v2=u2+2aS{v^2} = {u^2} + 2aS, where S is the total distance travelled by the particle, u is the initial velocity, v is the final velocity.