Question

A particle moves along a circle of radius \[\dfrac{{20}}{\pi }\;{\rm{m}}\] with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin the tangential acceleration is
A) $40\pi \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$
B) $160\pi \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$
C) $240\pi \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$
D) $640\pi \;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

Answer 1

Hint:Use the equation of motion in circular motion, \[\omega _f^2 = \omega _i^2 + 2\alpha \theta \] and the relationship between the angular and tangential acceleration is ${a_t} = \alpha r$.


Complete step by step solution:
We know from the question that the radius of the circular path in which the particle is moving is $r = \dfrac{{20}}{\pi }\;{\rm{m}}$, the final linear velocity of the particle at the end of the second revolution is ${v_f} = 80\;{\rm{m/s}}$.

Now we know that the particle travelled for two revolution of the circular path, so the angular displacement of the particle will be:
$
\theta = 2\pi + 2\pi \\
 = 4\pi \;{\rm{m}}
$

Now the final angular velocity of the particle at the end of two revolution is expressed as:
${\omega _f} = \dfrac{v}{r}$

Now we substitute $\dfrac{{20}}{\pi }\;{\rm{m}}$ as $r$ and $80\;{\rm{m/s}}$ as ${v_f}$ in the above expression.
\[
{\omega _f} = \dfrac{{80\;{\rm{m/s}}}}{{\left( {\dfrac{{20}}{\pi }\;{\rm{m}}} \right)}}\\
 = 4\pi \;{\rm{rad/s}}
\]

We have to assume that the particle started its motion from the rest, so the initial angular velocity of the particle is ${\omega _i} = 0$.

Now we use the equation of motion in circular motion to calculate the angular acceleration of the particle,
\[\omega _f^2 = \omega _i^2 + 2\alpha \theta \]
Here, $\alpha $ is angular acceleration of the particle.

Now we substitute \[4\pi \;{\rm{rad/s}}\] as ${\omega _f}$ , 0 as ${\omega _i}$ and $4\pi \;{\rm{m}}$ as $\theta $ in the above expression.
$
{\left( {4\pi } \right)^2} = 0 + 2\alpha \left( {4\pi } \right)\\
16{\pi ^2} = 8\pi \alpha \\
\alpha = 2\pi \;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}
$

Now we use the relationship between the tangential acceleration and angular acceleration,
${a_t} = \alpha r$

Now we substitute \[4\pi \;{\rm{rad/s}}\] as ${\omega _f}$ , 0 as ${\omega _i}$ and $4\pi \;{\rm{m}}$ as $\theta $ in the above expression.
$
{a_t} = 2\pi \times \dfrac{{20}}{\pi }\\
 = 40\;{\rm{m/s}}
$
Hence, the tangential acceleration of the particle after completing two revolutions is \[40\;\pi \;{\rm{m/}}{{\rm{s}}^2}\] and option (A) is correct.


Note:The tangential acceleration of a body is the result of change in speed of the body and it is expressed as ${a_t} = \dfrac{{dv}}{{dt}} = \alpha r$. Here particles travelled for two revolutions of the circular path.