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 #8 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Jul 19, 2017

Answer:

  1. #omega=8/3 1/s#
  2. #L=96pi (kg*m^2)/s#

Explanation:

For this equation, we need to find two things: 1. the angular velocity of the disk and 2. the angular momentum of the disk

Angular velocity, #omega#, is the rate of change of angular displacement, and it is the angular analogue to linear velocity. Likewise, these two share a relationship of #omega=v/r#
If we substitute our linear velocity of #8m/s# and our radius of #3m#into the equation:

#omega=v/r#

#omega=(8cancel(m)/s)/(3cancelm)#

#omega=8/3 1/s#

We now have the answer to part 1.

Part 2. is asking about angular momentum. One equation used to calculate this is #L=Iomega# where #L# is angular momentum, #I# is the moment of inertia, and #omega# is angular velocity.

The moment of inertia is an objects resistance to angular acceleration, and its calculation changes on the shape of the object. For a solid disk, the equation is
#I=1/2 Mr^2# where #M# is the object's mass and #r# is the radius of the disk. If we substitute in for those values, we see that

#I=1/2*4kg*3m^2#

#I=18kg*m^2#

If we substitute in this value into our first equation, we see that

#L=Iomega#

#L=18kg*m^2*(16pi)/3 1/s#

#L=96pi (kg*m^2)/s#