Question

A cat rides a merry-go-round turning with uniform circular motion. At time ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{s}$, the cat’s velocity is ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{=}\left(3.00m/s\right)\hat{\mathbf{i}}\mathbf{+}\left(4.00m/s\right)\hat{\mathbf{j}}$, measured on a horizontal xy coordinate system. At${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{s}$ , the cat’s velocity is${\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{j}}$ .What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s average acceleration during the time interval${\mathit{t}}_{\mathbf{2}}\mathbf{-}{\mathit{t}}_{\mathbf{1}}$, which is less than one period?

Short answer

1. The magnitude of the cat’s centripetal acceleration $a_c=5.24\mathrm{m}/{\mathrm{s}}^2$
2. The average acceleration of cat during the time interval$t_2-t_1,\mathrm{is}\mathrm{a}=3.33\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given data

The time interval, $t_1=2.00s$

The velocity of the cat at $t_1$is role="math" localid="1660902332545" $\overrightarrow{v_1}=\left(3.00\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(4.00\mathrm{m}/\mathrm{s}\right)\hat{j}$

The time interval, $t_2=5.00s$

The velocity of the cat at $t_2$ is $\overrightarrow{v_2}=\left(-3.00\mathrm{m}/\mathrm{s}\right)\hat{i}+\left(-4.00\mathrm{m}/\mathrm{s}\right)\hat{j}{t}_2-t_1<T$

Understanding the concept

Using the equation for centripetal acceleration, find the radius of the circular path. The particle is moving with uniform circular motion. Hence, the magnitude of velocity is constant. Sketch the figure as per the given condition where the particle covers half-quarters of the circular path. Find period T and then use the relation between velocity, acceleration, and radius and find the radius of the circular path and then the magnitude of centripetal acceleration. To find the average acceleration we can use the expression of acceleration in vector form and then find its magnitude.

Formula:

$$\begin{gathered} v=r\omega \\ \omega =\frac{2\tau \tau }{T} \\ {a}_c=\frac{v^2}{r} \\ \overrightarrow{a}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{t_2-t_1} \end{gathered}$$

Step 3: Sketch the figure as per the given conditions

(a) Calculate the magnitude of the cat’s centripetal acceleration

According to the figure, particle covers half quarters of the circumference of circle from $t_1$to$t_2$.

Hence,role="math" localid="1660907631624" $t_2-t_1$is equal to the half quarters of period T.

$$\begin{gathered} t_2-t_1=\frac{T}{2} \\ T=2\left(t_2-t_1\right) \\ =2\left(5.00s-2.00s\right) \\ =6.00s \\ {t}_2-t_1<T \end{gathered}$$

The particle is moving in circular path with uniform circular motion; hence the magnitude of velocity and acceleration remains the same. The magnitude of velocity is

$$\begin{gathered} v=\sqrt{{\left(v_1\right)}^2+{\left(v_2\right)}^2} \\ =\sqrt{{\left(3.00\mathrm{m}/\mathrm{s}\right)}^2+{\left(4.00\mathrm{m}/\mathrm{s}\right)}^2} \\ =5.00\mathrm{m}/\mathrm{s} \end{gathered}$$

The relation between velocity, acceleration and radius is,

$$\begin{gathered} v=r\omega \\ =r\frac{2\tau \tau }{T} \\ r=\frac{vT}{2\tau \tau } \\ =\frac{\left(5.00\mathrm{m}/\mathrm{s}\right)\times \left(6.00s\right)}{2\tau \tau } \\ =4.774\mathrm{m} \end{gathered}$$

The expression of the centripetal acceleration is,

$$\begin{gathered} a_c=\frac{v^2}{r} \\ =\frac{{\left(5.00\mathrm{m}/\mathrm{s}\right)}^2}{4.774\mathrm{m}} \\ =5.24\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The magnitude of the cat’s centripetal acceleration is $5.24\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculate the average acceleration  of cat during the time interval t2-t1

The expression of average acceleration is

$$\begin{gathered} \overrightarrow{\mathrm{a}}=\frac{\overrightarrow{{\mathrm{v}}_2}-\overrightarrow{{\mathrm{v}}_1}}{{\mathrm{t}}_2-{\mathrm{t}}_1} \\ =\frac{\left(-3.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(-4.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}-\left(4.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}}{5.00-2.00} \\ =\frac{\left(-6.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}+\left(-8.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}}{3.00} \\ =\left(-2.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}+\left(\frac{-8.00}{3.00}\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of the average acceleration is,

$$\begin{gathered} a=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2} \\ =\sqrt{{\left(-2.00\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(-8.00/3.00\mathrm{m}/{\mathrm{s}}^2\right)}^2} \\ =3.33\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of acceleration is $3.33\mathrm{m}/{\mathrm{s}}^2$.