A cylinder has inner and outer radii of #8 cm# and #12 cm#, respectively, and a mass of #9 kg#. If the cylinder's frequency of counterclockwise rotation about its center changes from #1 Hz# to #7 Hz#, by how much does its angular momentum change?

1 Answer
Apr 6, 2016

Answer:

#DeltaL= 9/2(8.0xx10^-3)=0.432pi~~1.357(kgm^2)/s#

Explanation:

Given:
1) A hollow cylinder radii: #R_o=8cm; R_i=12cm#
2) Mass, #m=12 kg #
3) #omega_1 =2pif_1; omega_2 =2pif_2; f_1=1Hz; f_2=7Hz#

Required change in angular momentum:
#DeltaL= I(omega_2-omega_1)#

Theorem, Definition and Principles:
Angular Momentum, #L=Iomega# where
#I# = Moment of Inertia
#omega# = Angular Velocity

Solution Strategy:
A) Compute the Moment of Inertia of Hollow cylinder
B) Convert frequency to radians: #omega =2pif#
C) Compute the Angular momentum for #omega_1 and omega_2#

#color(brown)("A)")# Moment of Inertia of Hollow Cylinder (HC)
Let #R_o=8cm; R_i=12cm# be outer and inner radii of HC
then we can derive the moment if inertia, I of HC as follows:
#I= intr^2dm# where #r# is arbitrary radius in #r:r in[R_o,R_i]#
Also let #rho=M/V# be the density and #l="length of cylinder"#
then #dm=rhocolor(blue)(dv)=rho*color(blue)(2pi*r*l*dr)#
Now because we are dealing with HC,
#rho = M/(pi(R_i^2-R_o^2)l) #

Integrate over the inner and outer radiuses and inserting the expression for #rho#
#I=rho2pil int_(R_o)^(R_i) r^3dr#
#= M/(pi(R_i^2-R_o^2)l) 2pil 1/4[R_i^4-R_o^4 ]#

#= M/(cancel(pi(R_i^2-R_o^2))l) 2l 1/4cancel(pi[R_i^2-R_o^2 ])[R_i^2+R_o^2]#
#I_(HC_0)=1/2M [R_i^2+R_o^2]#

#color(brown)("B) Convert "omega)#
#omega_1= 2pi; omega_2=2pi*7=14pi#

#color(brown)("C) Compute the angular momentum before and after"#
#L_o=Iomega_1 # Before
#L_i=Iomega_2 # After
Change #DeltaL= I(omega_2-omega_1)#
#=9/2[1.44xx10^-2-6.4xx10^-3]2pi(7-1)#
#DeltaL= 9/2(8.0xx10^-3)=0.432pi~~1.357(kgm^2)/s#