What is the angular momentum of an object with a mass of $5 k g$ $5 kg$ that moves along a circular path of radius $9 m$ $9 m$ at a frequency of $\frac{2}{3} H z$ $2/3 Hz$ ?

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What is the angular momentum of an object with a mass of $5 k g$ $5 kg$ that moves along a circular path of radius $9 m$ $9 m$ at a frequency of $\frac{2}{3} H z$ $2/3 Hz$ ?

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Answer 1 · 2015-12-17T17:45:07

We will first convert $H z$ $Hz$ into $\frac{r a d}{sec}$ ${rad}/sec$ using the formula below:
$ω = 2 π f$ $\omega=2\pi f$
$ω = 2 \times π \times \frac{2}{3}$ $\omega= 2\times \pi\times2/3$
$ω = 4.188 \frac{r a d}{sec}$ $\omega=4.188{rad}/sec$

We then use the formula for Angular momentum
$L = \to r \times \to p$ $L=\vec{r}\times \vec{p}$ = $L = \to r \times m \to v$ $L=\vec{r}\times m \vec{v}$
Since we are calculating Angular momentum wrt to the center of the circle, and during circular motion $\to v$ $\vec{v}$ is perpendicular to $\to r$ $\vec{r}$ hence the cross product simply becomes a simple multiplication, with the angular momentum pointing along Z-axis( assuming motion is happening in the X-Y plane)

Velocity in a circular motion, given its angular velocity is given by:
$v = ω r$ $v=\omega r$ , where $r$ $r$ is the radius of circle

Therefore,
$L = m ω r^{2} ˆ z$ $L=m\omegar^2 \hat{z}$
$L = 5 \times 4.188 \times 9^{2}$ $L= 5\times 4.188\times 9^2$ = $1696.46 K g$ $1696.46 Kg$ $\frac{m^{2}}{sec}$ $m^2/sec$ $ˆ z$ $\hat{z}$

Answer 2 · 2015-12-17T17:48:43

$L = 1696.7 {kg.m}^{2} s^{- 1}$ $L= 1696.7" ""kg.m"^2"s"^(-1)$

Explanation:

The expression for angular momentum is:

$L = m \times v \times r$ $L=mxxvxxr$

$m$ $m$ = mass

$v$ $v$ = velocity

$r$ $r$ = radius

We can use the information given to find the velocity $v$ $v$ :

The frequency of rotation $f$ $f$ is $\frac{2}{3} Hz$ $2/3"Hz"$

This means that the time period $T$ $T$ is $\frac{1}{f}$ $1/f$ which equals $\frac{3}{2} = 1.5 s$ $3/2=1.5"s"$

This means that it takes $1.5 s$ $1.5"s"$ to make one complete revolution.

We know the circumference of the circle is $2 π r$ $2pir$ .

This is the distance travelled in $1.5 s$ $1.5"s"$

So the velocity is given by:

$v = \frac{2 π 9}{1.5} = 12 π m/s$ $v=(2pi9)/1.5=12pi"m/s"$

So we can now get $L$ $L$ :

$L = 5 \times 12 π \times 9$ $L=5xx12pixx9$

$:.L= 1696.7" ""kg.m"^2"s"^(-1)$

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What is the angular momentum of an object with a mass of $5 k g$ $5 kg$ that moves along a circular path of radius $9 m$ $9 m$ at a frequency of $\frac{2}{3} H z$ $2/3 Hz$ ?

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