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Question: A particle of mass 10g moves along a circle of radius \(6.4cm\) with a constant tangential accelerat...

A particle of mass 10g moves along a circle of radius 6.4cm6.4cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10−4J8 \times {10^{ - 4}}J by the end of the second revolution after the beginning of the motion?
(A) 0.2m/s20.2m/{s^2}
(B) 0.1m/s20.1m/{s^2}
(C) 0.15m/s20.15m/{s^2}
(D) 0.18m/s20.18m/{s^2}

Explanation

Solution

We know that the kinetic energy that a body possesses due to its motion. It is dependent on the motion and its speed. The kinetic energy is gained by the body due to acceleration and changes according to its speed. Mathematically the kinetic energy will be given by,
K=12mv2K = \dfrac{1}{2}m{v^2}.

Formula used: We will be using the formula K=12mv2K = \dfrac{1}{2}m{v^2} to find the kinetic energy of a body in motion, where KK is the kinetic energy possessed by the body, mm is the mass of the body in motion, vv is the velocity at which the body is in motion.
We will also be using the formula v2−u2=2as{v^2} - {u^2} = 2as where vv is the final velocity achieved by the body, uu is the initial velocity of the body, aa is the acceleration experienced by the body, ss is the displacement of the body.

Complete Step by Step answer:
We know that kinetic energy is the form of energy that a body achieves by the virtue of its motion. That states that a body at rest has no kinetic energy and the body in motion due to its acceleration has some kinetic energy stored in the body.
From the question we know that the particle has a mass of m=10gm = 10g and moves along a circle of radius r=6.4cmr = 6.4cm . The kinetic energy of the body during the second revolution is K=8×10−4JK = 8 \times {10^{ - 4}}J .
We know that Kinetic energy is given by, K=12mv2K = \dfrac{1}{2}m{v^2}
8×10−4J=12mv28 \times {10^{ - 4}}J = \dfrac{1}{2}m{v^2}
Substituting the values of m=10g=10×10−3kgm = 10g = 10 \times {10^{ - 3}}kg
(10×10−3kg)v2=8×10−4×2\left( {10 \times {{10}^{ - 3}}kg} \right){v^2} = 8 \times {10^{ - 4}} \times 2
v2=8×10−4×2(10×10−3){v^2} = \dfrac{{8 \times {{10}^{ - 4}} \times 2}}{{\left( {10 \times {{10}^{ - 3}}} \right)}}
v2=16×10−4(10−2){v^2} = \dfrac{{16 \times {{10}^{ - 4}}}}{{\left( {{{10}^{ - 2}}} \right)}}
v2=16×10−2{v^2} = 16 \times {10^{ - 2}}
⇒v=4×10−1m/s\Rightarrow v = 4 \times {10^{ - 1}}m/s
Thus, we now know that the velocity of the body during its second revolution will be v=4×10−1m/sv = 4 \times {10^{ - 1}}m/s .
We also know from the laws of kinematics that v2−u2=2as{v^2} - {u^2} = 2as ,
Here let us assume that the body starts from rest, u=0u = 0 . We know that the body undergoes circular motion and the distance travelled after 2 revolutions would be, s=2(2πr)s = 2\left( {2\pi r} \right) .
However, the radius of the circle r=6.4cm=6.4×10−2mr = 6.4cm = 6.4 \times {10^{ - 2}}m .
Substituting in the laws of kinematics we get,
(4×10−1m/s)2−(0)2=2a(4π(6.4×10−2m)){\left( {4 \times {{10}^{ - 1}}m/s} \right)^2} - {\left( 0 \right)^2} = 2a\left( {4\pi \left( {6.4 \times {{10}^{ - 2}}m} \right)} \right)
Solving for aa we get,
a=(4×10−1m/s)22(4π(6.4×10−2m))a = \dfrac{{{{\left( {4 \times {{10}^{ - 1}}m/s} \right)}^2}}}{{2\left( {4\pi \left( {6.4 \times {{10}^{ - 2}}m} \right)} \right)}}
Substituting the value of π=3.14\pi = 3.14
a=16×10−28×3.14×6.4×10−2a = \dfrac{{16 \times {{10}^{ - 2}}}}{{8 \times 3.14 \times 6.4 \times {{10}^{ - 2}}}}
Solving the arithmetic’s, we get, a=0.099522m/s2a = 0.099522m/{s^2}
Thus, the acceleration of the body after the second revolution will be a=0.099522m/s2≈0.1m/s2a = 0.099522m/{s^2} \approx 0.1m/{s^2}

So the correct answer will be option B.

Note: We know that the body when in motion during acceleration has kinetic energy. What happens when the body is decelerating towards rest? The body still has the same kinetic energy in the negative direction numerically. This kinetic energy will be stored as the potential energy of the body once it is in rest. Because energy in any system is always conserved.