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

Answer:

#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)#