Question

A particle of mass $10kg$ moves along a circle of radius $6.4cm$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8\times {10}^{-4}J$ at the end of the second revolution after the beginning of the motion?

Answer 1

Step 1: Given data

A particle of mass $m=10kg$ and moves along a circle of raius $r=6.4cm$.

The kinetic energy of the body during the second revolution $k=8\times {10}^{-4}J$

Step 2: Formula used

$K=\frac{1}{2}m{v}^2{[whereK=kineticenergy,m=mass,v=velocity]}^{}$

$v^2=u^2+2as[wherev=finalvelocity,u=intialvelocity,a=acceleration,s=displacement]$

Step 3: Calculating velocity

Kinetic energy is the type of energy that a body generates as a result of its movement. That is, a body at rest has no kinetic energy, whereas a body in motion has some kinetic energy stored in it due to its acceleration.

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ {v}^2=\frac{K\times 2}{m} \end{gathered}$$

Substituting the given value of K and m

$$\begin{gathered} v^2=\frac{8\times {10}^{-4}\times 2}{(10\times {10}^{-3})}[As10g=10\times {10}^{-3}kg] \\ {v}^2=\frac{16\times {10}^{-4}}{{10}^{-2}} \\ {v}^2=16\times {10}^{-2} \\ v=4\times {10}^{-1}m/s \end{gathered}$$

Thus, we now know that the velocity of the body during its second revolution will be

Step 4: Calculating acceleration

Here let us assume that the body starts from rest, $u=0$. We know that the body undergoes circular motion and the distance traveled after 2 revolutions would be, $s=2\left(2\pi r\right)$.

However, the radius of the circle$r=6.4cm=6.4\times {10}^{-2}m.$

According to Newton's third law of motion.

$$\begin{gathered} v^2=u^2+2as \\ {v}^2-u^2=2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}$$

Substituting the value of v, u, and s.

$$\begin{gathered} a=\frac{{(4\times {10}^{-1}m/s)}^2}{2\left(4\pi \right(6.4\times {10}^{-2}m\left)\right)} \\ a=\frac{16\times {10}^{-2}}{8\times 3.14\times 6.4\times {10}^{-2}}[As,\pi =3.14] \\ a=0.099522m/s^2 \\ a\approx 0.1m/s^2 \end{gathered}$$

Hence, the acceleration of the body after the second revolution will be $0.1m/s^2$.