What is the angular momentum of an object with a mass of 5 kg that moves along a circular path of radius 9 m at a frequency of 2/3 Hz ?

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2 Answers
Dec 17, 2015

We will first convert Hz into {rad}/sec using the formula below:
\omega=2\pi f
\omega= 2\times \pi\times2/3
\omega=4.188{rad}/sec

We then use the formula for Angular momentum
L=\vec{r}\times \vec{p}= L=\vec{r}\times m \vec{v}
Since we are calculating Angular momentum wrt to the center of the circle, and during circular motion \vec{v} is perpendicular to \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=\omega r, where r is the radius of circle

Therefore,
L=m\omegar^2 \hat{z}
L= 5\times 4.188\times 9^2= 1696.46 Kg m^2/sec \hat{z}

Dec 17, 2015

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

Explanation:

The expression for angular momentum is:

L=mxxvxxr

m = mass

v = velocity

r = radius

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

The frequency of rotation f is 2/3"Hz"

This means that the time period T is 1/f which equals 3/2=1.5"s"

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

We know the circumference of the circle is 2pir.

This is the distance travelled in 1.5"s"

So the velocity is given by:

v=(2pi9)/1.5=12pi"m/s"

So we can now get L:

L=5xx12pixx9

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