Question

A flea is at point (A) on a horizontal turntable$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{m}}$,from the center. The turntable is rotating at$\mathbf{33}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{min}$in the clockwise direction. The flea jumps straight up to a height of$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{cm}}$. At takeoff, it gives itself no horizontal velocity relative to the turntable. The flea lands on the turntable at point (B). Choose the origin of coordinates to be at the center of the turntable and the positive$\mathit{x}$axis passing through (A) at the moment of takeoff. Then the original position of the flea is$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\hat{\mathbf{i}}\mathbf{ }\mathbf{cm}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\left(n-r\right)\mathbf{!}}$.

(a) Find the position of point (A) when the flea lands.

(b) Find the position of point (B) when the flea lands.

Short answer

(a) The position of point A when the flea land is$\left(\right(7.62) \hat{\mathrm{i}}+(-6.48) \hat{\mathrm{j}} ) \mathrm{cm}$.

(b) The position of point when the flea land is $(10.0\hat{\mathrm{i}}-7.05\hat{\mathrm{j}}) \mathrm{cm}$.

Step-by-step solution

Understanding the concept:

The formula to calculate the maximum height is,

$\mathit{H}\mathbf{=}\frac{{\left(v_i\right)}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{2}\mathbf{g}}$

Here,

$\mathbf{g}$is the acceleration due to gravity,$\mathbf{H}$is the maximum height reached by the flea, ${\mathbf{V}}_{\mathbf{i}}$is the initial velocity, $\mathbf{\theta }$is the angle with horizontal.

The formula to calculate the time taken by the flea is,

$\mathit{t}\mathbf{=}\frac{\mathbf{2}{\mathbf{v}}_{\mathbf{i}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{g}}$

Here,$t$is the time of flight of flea.

The number of rotations made by turn table is,

role="math" localid="1663612189743" $\mathit{\theta }\mathbf{=}\mathit{\omega }\mathit{t}$

Here,$\mathit{\omega }$is the angular velocity,$\mathit{\theta }$is the number of rotation.

The formula to calculate the linear velocity of the flea is,

$\mathit{v}\mathbf{=}\mathit{d}\mathit{\omega }$

Here,$\mathit{v}$is the linear velocity.

Determine the position of the point (A) when the flea lands.

(a)

Given info: The distance of flea from the center of the table is $10.0\text{cm}$. The angular speed of the turn table is $33.3\mathrm{rev}/\min$in clockwise direction, the height of the flea after jump is $5.00\text{cm}$, at the time of takeoff the horizontal speed of the flea relative to the table is zero.

The original distance of the flea from the center of the table is $\overrightarrow{\mathrm{d}}=10.0 \hat{\mathrm{i}} \mathrm{cm}$,

Rearrange the formula to calculate the maximum height

$\begin{array}{c}{\left(v_{\mathrm{i}}\right)}^2{\sin }^2\theta =2gH\\ v_{\mathrm{i}}\sin \theta =\sqrt{2gH}\end{array}$

Substitute $5.00\text{cm}$for $H$and $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in the above equation.

$\begin{array}{c}{v}_{\mathrm{i}}\sin \theta =\sqrt{2(5.00\mathrm{cm})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =\sqrt{2(5.00\mathrm{cm})\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =0.98\mathrm{m}/\mathrm{s}\end{array}$

Thus, the vertical component of the initial velocity is $=0.98\mathrm{m}/\mathrm{s}$.

The formula to calculate the time taken by the flea is,

$t=\frac{2v_i\sin \theta }{g}$

Substitute $0.98\mathrm{m}/\mathrm{s}$for $v_i\sin \theta$and $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in the above equation.

$\begin{array}{c}t=\frac{2(0.98\mathrm{m}/\mathrm{s})}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =0.20\mathrm{s}\end{array}$

Thus, the time of flight for flea is $0.20\text{s}$.

The number of rotations made by turn table is,

$\theta =\omega t$

Substitute localid="1663613688841" $33.3\mathrm{rev}/\min$for $\omega$and $0.20\text{s}$for $t$in the above equation.

$\begin{array}{c}\theta =(33.3\mathrm{rev}/\min )(0.20\mathrm{s})\\ =(33.3\mathrm{rev}/\min )\left(\frac{{360}^{\circ }}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)(0.20\mathrm{s})\\ =40.3°\end{array}$

The position of A when flea land along $x$axis is,

$x=d\cos \theta \hat{\mathrm{i}}$

Substitute $10.0\text{cm}$for $d$and $40.3°$for $\theta$in the above equation.

$\begin{array}{c}x=(10.0\mathrm{cm})\cos \left(40.3°\right)\\ =7.62\mathrm{cm}\end{array}$

Thus, the position of A along axis is $7.62\text{cm}$.

The position of point A when flea land along $y$axis is,

$y=d\sin \theta$

Substitute $10.0\text{cm}$for $d$and $40.3°$for $\theta$in the above equation.

$\begin{array}{c}y=-(10.0\mathrm{cm})\sin \left(40.3°\right)\\ =-6.48\mathrm{cm}\end{array}$

Thus, the position of A along $y$axis is$-6.48\text{cm}$.

The position coordinate of A is,

$\overrightarrow{\mathrm{r}}=(x\hat{\mathrm{i}}+y\hat{\mathrm{j}})$

Substitute $7.62\text{cm}$for $x$and $-6.48\text{cm}$for $y$in the above equation.

$\overrightarrow{\mathrm{r}}=\left(\right(7.62)\hat{\mathrm{i}}+(-6.48\left)\hat{\mathrm{j}}\right) \mathrm{cm}$

Therefore, the position of point A when the flea land is $\left(\right(7.62)\hat{\mathrm{i}}+(-6.48\left)\hat{\mathrm{j}}\right) \mathrm{cm}$.

Determine the position of the point (B) when the flea lands.

(b)

The formula to calculate the linear velocity of the flea is,

$v=d\omega$

Substitute$33.3\mathrm{rev}/\min$for$\omega$and$10.0\text{cm}$for$t$in the above equation.

role="math" localid="1663614059305" $\begin{array}{c}v=(10.0\mathrm{cm})(33.3\mathrm{rev}/\min )\\ =(10.0\mathrm{cm})\left(\frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}\right)(33.3\mathrm{rev}/\min )\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)\\ \approx 0.3525\mathrm{m}/\mathrm{s}\end{array}$

Thus, the linear velocity of the flea is$0.3525\mathrm{m}/\mathrm{s}$.

From part (a) time of flight of flea is$0.20\text{s}$.

The distance travelled by flea alongaxis is,

$y^{\text{'}}=vt\hat{j}$

Substitute $0.3525\mathrm{m}/\mathrm{s}$for $d$and $0.20\text{s}$for $t$in the above equation.

$\begin{array}{c}{y}^{\text{'}}=(0.3525\mathrm{m}/\mathrm{s})(0.20\mathrm{s})\hat{\mathrm{j}}\\ =0.0705\hat{\mathrm{j}} \mathrm{m}\left(\frac{{10}^2\mathrm{cm}}{1\mathrm{m}}\right)\\ =7.05\hat{\mathrm{j}} \mathrm{cm}\end{array}$

The position is below$x$axis.

Thus, the position of B along$y$axis is$-7.05\hat{j}\mathrm{cm}$.

The position coordinate ofis,

${\overrightarrow{r}}_{\mathrm{B}}=d\hat{\mathrm{i}}+y^{\text{'}}\hat{\mathrm{j}}$

Here,

${\overrightarrow{r}}_{\mathrm{B}}$is the position of point.

Substitute$10.0\text{cm}$for$d$and$-7.05\text{cm}$for$y\text{'}$in the above equation.

$\begin{array}{c}{\overrightarrow{r}}_{\mathrm{B}}=(10.0\mathrm{cm})\hat{\mathrm{i}}+(-7.05\mathrm{cm})\hat{\mathrm{j}}\\ =(10.0\hat{\mathrm{i}}-7.05\hat{\mathrm{j}}) \mathrm{cm}\end{array}$

Therefore, the position of point when the flea land is $(10.0\hat{\mathrm{i}}-7.05\hat{\mathrm{j}}) \mathrm{cm}$.