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# 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 $8 \times {10^{ - 4}}J$ by the end of the second revolution after the beginning of the motion?A. $0.1{\text{ m/}}{{\text{s}}^2}$B. $0.15{\text{ m/}}{{\text{s}}^2}$C. $0.18{\text{ m/}}{{\text{s}}^2}$D. $0.2{\text{ m/}}{{\text{s}}^2}$

Last updated date: 02nd Aug 2024
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Hint: 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 $m = 10g = 10 \times {10^{ - 3}}kg$
The radius of the circle $r = 6.4cm = 6.4 \times {10^{ - 2}}m$
Kinetic energy =$8 \times {10^{ - 4}}J$
Where kinetic energy is given by the formula$KE = \dfrac{1}{2}m{v^2} - - (i)$
Hence we can write
$\dfrac{1}{2}m{v^2} = 8 \times {10^{ - 4}}J \\ \dfrac{1}{2}\left( {10 \times {{10}^{ - 3}}} \right){v^2} = 8 \times {10^{ - 4}}J \\$
By further solving we get
${v^2} = \dfrac{{2 \times 8 \times {{10}^{ - 4}}}}{{10 \times {{10}^{ - 3}}}} \\ {v^2} = 16 \times {10^{ - 2}} \\ v = 4 \times {10^{ - 1}} \\ v = 0.4{\text{ m/s}} - - (ii) \\$
Now find the total distance covered by the particle after two revolutions by using
$S = 2\left( {2\pi r} \right) - - (iii)$
By solving we get
$S = 4\pi r \\ = 4 \times 3.14 \times 6.4 \times {10^{ - 2}} \\ = 0.8m \\$
Now use the third equation of newton’s law of motion which is given as
${v^2} = {u^2} + 2aS - - (iii)$
Now since the particle started from the rest, hence its initial velocity will be
$u = 0$
Now substitute the value of distance S and the velocity v in equation (iii) to find the acceleration; hence we get
${v^2} = {u^2} + 2aS - - (iii) \\ {\left( {0.4} \right)^2} = 0 + 2a(0.8) \\ 0.16 = 2a(0.8) \\ a = 0.1{\text{ m/}}{{\text{s}}^2} \\$
Therefore, the magnitude of this acceleration by the end of the second revolution $= 0.1{\text{ m/}}{{\text{s}}^2}$
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 ${v^2} = {u^2} + 2aS$, where S is the total distance travelled by the particle, u is the initial velocity, v is the final velocity.