Question

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. \[0.2{\text{ m/}}{{\text{s}}^2}\]

Answer 1

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.