Question

: A cat rides a merry-go-round turning with uniform circular motion. At time ${t_1} = 2.00{\rm{ s}}$, the cat's velocity is ${\vec v_1} = \left( {3.00{\rm{ m/s}}} \right)\hat i + \left( {4.00{\rm{ m/s}}} \right)\hat j$, measured on a horizontal $xy$ coordinate system. At ${t_2} = 5.00{\rm{ s}}$, the cat's velocity is ${\vec v_2} = \left( { - 3.00{\rm{ m/s}}} \right)\hat i + \left( { - 4.00{\rm{ m/s}}} \right)\hat j$. What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval ${t_2} - {t_1}$, which is less than one period?

Answer 1

This question is based on the relation between rectilinear motion and the circular motion of a particle. If the velocity vector of a particle is given by $\vec v = a\hat i + b\hat j$, then the magnitude of the velocity is given by,
$\left| {\vec v} \right| = \sqrt {{a^2} + {b^2}}$

Complete step by step answer:
Given:
The initial time period ${t_1} = 2.00{\rm{ s}}$
At this time, the cat’s velocity ${\vec v_1} = \left( {3.00{\rm{ m/s}}} \right)\hat i + \left( {4.00{\rm{ m/s}}} \right)\hat j$
The final time period ${t_2} = 5.00{\rm{ s}}$
At this time, the cat’s velocity ${\vec v_2} = \left( { - 3.00{\rm{ m/s}}} \right)\hat i + \left( {4.00{\rm{ m/s}}} \right)\hat j$
So, total time elapsed is $\begin{array}{l} \left( {{t_2} - {t_1}} \right) = 5.00 - 2.00\\\ = 3.00{\rm{ s}} \end{array}$
As we know that the cat is moving in the circular motion, so it takes $3.00{\rm{ s}}$ to move to the opposite side of the circle, so total time taken to complete one cycle of the circular motion is given by,
$\begin{array}{l} T = 2 \times 3.00{\rm{ s}}\\\ T = 6.00{\rm{ s}} \end{array}$
And we know that the angular velocity of the cat $\omega = \dfrac{{2\pi }}{T}$
The relation between the linear velocity $v$ and the angular velocity is given by,
$v = \omega \times r$
Where, $r$ is the radius of the circle.
To find the magnitude of linear velocity $v$ we use the velocity vector ${\vec v_1}$ in the following way,
The velocity can be calculated as,
$\begin{array}{l} v = \sqrt {{{\left( {3.00} \right)}^2} + {{\left( {4.00} \right)}^2}} \\\ \Rightarrow v = \sqrt {9 + 16} \\\ \Rightarrow v = \sqrt {25} \\\ \Rightarrow v = 5{\rm{ m/s}} \end{array}$
So now, we substitute the value of $v$ in the expression $v = \omega r$ and calculate the value of the radius of the circle $r$.
$\begin{array}{l} \Rightarrow r = \dfrac{v}{\omega }\\\ \Rightarrow r = \dfrac{5}{{\left( {\dfrac{{2\pi }}{T}} \right)}} \end{array}$
Substitute the value of $T = 6.00{\rm{ s}}$ we get,
$\begin{array}{l} r = \dfrac{{5 \times 6.00}}{{2\pi }}\\\ r = 4.77{\rm{ m}} \end{array}$
(a)The magnitude of the cat’s centripetal acceleration ${a_c} = \dfrac{{{v^2}}}{r}$
Substituting the values, we get,
$\begin{array}{l} {a_c} = \dfrac{{{5^2}}}{{4.77}}\\\ \Rightarrow {a_c} = 5.236\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}} \end{array}$
Therefore, the magnitude of the cat’s centripetal acceleration is $5.236{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}$.
(b) The cat’s average acceleration is given by the formula,
${\rm{Average acceleration = }}\dfrac{{{\text{Change in velocity}}}}{{{\text{Elapsed time}}}}$
Change in velocity \begin{array}{l} \left( {{{\vec v}_2} - {{\vec v}_1}} \right) = \left\\{ {\left( { - 3.00{\rm{ m/s}}} \right)\hat i + \left( { - 4.00{\rm{ m/s}}} \right)\hat j} \right\\} - \left\\{ {\left( {3.00{\rm{ m/s}}} \right)\hat i + \left( {4.00{\rm{ m/s}}} \right)\hat j} \right\\}\\\ = \left( { - 3.00 - 3.00} \right)\hat i + \left( { - 4.00 - 4.00{\rm{ }}} \right)\hat j\\\ = \left( { - 6.00} \right)\hat i + \left( { - 8.00} \right)\hat j \end{array}
And time period
$\begin{array}{l} \left( {{t_2} - {t_1}} \right) = 5.00 - 2.00\\\ = 3.00{\rm{ s}} \end{array}$
So, putting these values in the formula, we get,
Average acceleration
$\begin{array}{l} {{\vec a}_{avg}} = \dfrac{{\left( {{{\vec v}_2} - {{\vec v}_1}} \right)}}{{\left( {{t_2} - {t_1}} \right)}}\\\ {{\vec a}_{avg}} = \dfrac{{\left( { - 6.00} \right)\hat i + \left( { - 8.00} \right)\hat j}}{{3.00}}\\\ {{\vec a}_{avg}} = \left( { - 2.00} \right)\hat i + \left( { - 2.67} \right)\hat j \end{array}$
The magnitude of the average acceleration is given by,

\left| {{{\vec a}_{avg}}} \right| = \sqrt {{{\left( { - 2.00} \right)}^2} + {{\left( { - 2.67} \right)}^2}} \\\ \left| {{{\vec a}_{avg}}} \right| = 3.33{\rm{ }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}} \end{array}$$ Therefore, the cat’s average acceleration is ${\rm{3}}{\rm{.33 }}{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}$. **Note:** It should be noted that velocity vectors ${\vec v_1}{\rm{ and }}{\vec v_2}$ have same magnitude, but the direction of both of the vectors is opposite to each other and for a particle in a circular motion, it is only possible when the particle reaches the opposite side of the circle. It also means that the particle has completed half of the cycle at this point.