A solid disk, spinning counter-clockwise, has a mass of #11 kg# and a radius of #4/7 m#. If a point on the edge of the disk is moving at #8/5 m/s# in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Dec 26, 2016

Answer:

The angular momentum of the disk is #~~5# #(kgm^2)/s# and the angular velocity is #14/5 (rad)/s (~~2.8 (rad)/s)#.

Explanation:

Angular momentum is given by #vecL=Iomega#, where #I# is the moment of inertia of the object, and #omega# is the angular velocity of the object.

The moment of inertia of a solid disk is given by #I=1/2mr^2#, and angular velocity is given by #v/r#, where #v# is the tangential velocity and #r# is the radius.

We are given that #m=11kg#, #r=4/7m#, and #v=8/5m/s#. We can use these values to calculate the moment of inertia and angular velocity, and ultimately the angular momentum.

#omega=v/r=(8/5m/s)/(4/7m)#

#=>omega=14/5(rad)/s#

This is the angular velocity.

#I=1/2mr^2=1/2(11kg)(4/7m)^2#

#=>I=88/49 kgm^2#

This is the moment of inertia.

We can now calculate the angular momentum:

#vecL=Iomega=(88/49kgm^2)*(14/5(rad)/s)#

#=>vecL=176/35 (kgm^2)/s#

#=>vecL~~5 (kgm^2)/s#