A solid disk, spinning counter-clockwise, has a mass of #4 kg# and a radius of #3 m#. If a point on the edge of the disk is moving at #2/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Question

A solid disk, spinning counter-clockwise, has a mass of #4 kg# and a radius of #3 m#.  If a point on the edge of the disk is moving at #2/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

Benjamin Asiedu · Accepted Answer

The angular momentum is #=2.4kgm^2s^-1#
The angular velocity is #=0.13rads^-1# The velocity at an angle is #omega=(Deltatheta)/(Deltat)# #v=r*((Deltatheta)/(Deltat))=r omega# #omega=v/r# where, #v=2/5ms^(-1)# #r=3m# So, The angular velocity is #omega=(2/5)/(3)=0.13rads^-1# The angular momentum is #L=Iomega# where #I# is the moment of inertia For a solid disc, #I=(mr^2)/2# So, #I=4*(3)^2/2=18kkgm^2# The momentum of angular motion is #L=0.13*18=2.4kgm^2s^-1#

Xavier Adame · Answer

undefined Angular momentum: $ L = I \omega $, where $ I $ is the moment of inertia and $ \omega $ is the angular velocity.

Moment of inertia for a solid disk: $ I = \frac{1}{2} m r^2 $.

Angular velocity: $ \omega = \frac{v}{r} $, where $ v $ is the linear velocity and $ r $ is the radius.

Calculate moment of inertia ($ I $).

Calculate angular velocity ($ \omega $) using $ \omega = \frac{v}{r} $.

Calculate angular momentum ($ L $) using $ L = I \omega $.

Linear velocity ($ v $) is given as $ \frac{2}{5} $ m/s.