A solid disk, spinning counter-clockwise, has a mass of #2 kg# and a radius of #5/2 m#. If a point on the edge of the disk is moving at #15/4 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
A08
Dec 19, 2016

Angular momentum #=9.375kg cdot m^2s^-1#
Angular velocity#=1.5radcdot s^-1#

Explanation:

For any object which is rotating about an axis, each point located on the object has the same angular velocity #omega#. Its units are #radcdot s^-1#. It can be found with the help of velocity #v# and radius #r# of a point using the relation

#omega=v/r#

Inserting given values we get
#omega=(15/4)/(5/2)#
#=>omega=(15/4)xx(2/5)=1.5radcdot s^-1#

The angular momentum #L# of a solid disc can be found using the following formula

#L = Ixxω#
where #I# is the moment of inertia of a solid disc and is given as #1/2Mr^2#. #M# being mass of disc.

If we substitute the value of #omega# in terms of #vand r#, the expression for angular momentum reduces to
#L = 1/2Mr^2xxv/r#
#=>L = 1/2Mrv#
Inserting given values
#L = 1/2xx2xx5/2xx15/4#
#=>L = 9.375kg cdot m^2s^-1#

Related questions
See all questions in Angular Momentum
Impact of this question
1434 views around the world
You can reuse this answer
Creative Commons License