A cylinder has inner and outer radii of #2 cm# and #5 cm#, respectively, and a mass of #5 kg#. If the cylinder's frequency of counterclockwise rotation about its center changes from #12 Hz# to #9 Hz#, by how much does its angular momentum change?

2 Answers
Apr 28, 2018

Let's assume this cylinder is uniform (which is very important).

Recall for a solid cylinder,

#I = 1/2mr^2#

Moreover, recall angular momentum,

#L = Iomega#

Now, let's derive the moment of inertia of this circular body,

#I = 1/2mr^2 approx 5"kg"*"m"^2#

which is generally constant in our calculations.

Now, recall,

#omega = 2pif#

Hence,

#I_(1)2pif_1 approx (75"kg"*"m"^2)/"s" #, and

#I_(2)2pif_2 approx (57"kg"*"m"^2)/"s"#

#therefore DeltaL = L_2 - L_1 approx (18"kg"*"m"^2)/"s"#

is the change in angular momentum given your data.

Apr 29, 2018

Answer:

The change in angular momentum is #=0.137kgm^2s^-1#

Explanation:

The angular momentum is #L=Iomega#

where #I# is the moment of inertia

and #omega# is the angular velocity

The mass of the cylinder is #m=5kg#

The radii of the cylinder are #r_1=0.02m# and #r_2=0.05m#

For the cylinder, the moment of inertia is #I=m((r_1^2+r_2^2))/2#

So, #I=5*((0.02^2+0.05^2))/2=0.00725kgm^2#

The change in angular velocity is

#Delta omega=Deltaf*2pi=(12-9) xx2pi=6pirads^-1#

The change in angular momentum is

#DeltaL=IDelta omega=0.00725xx6pi=0.137kgm^2s^-1#