Question

A flat uniform circular disk (radius = 2.00 m, mass = 1.00

x102 kg) is initially stationary. The disk is freeto
rotate in the horizontal plane about a frictionless
axisperpendicular to the center of the disk. A 40.0 kg
person,standing 1.25 m from the axis, begins to run on the disk in
acircular path and has a tangential speed of 2.00 m/s relative
tothe ground. Find the resulting angular speed of the disk
(inrad/s) and describe the direction of the rotation.

Answer 1

The resulting angular speed of the disk is 0.5 rad/s

Step-by-step explanation:

Step 1: Data given

Radius of the circular disk = 2.00 meters

Mass of the circular disk = 1.00

Mass op the person = 40.0 kg

Distance from the axis = 1.25 m

tangential speed = 2.00 m/s

Step 2:

There is no external torque acting on the system so we can apply the law of conservation of angular momentum In this case the momentum is conserved.

Angular momentum of the man = Iω

⇒ With I = Inertia of the man about the axis of rotation = M*r²

⇒ I = 40 *1.25² = 62.5

⇒ with ω = Angular velocity of the man

⇒ v = 2m/s

⇒ Circumference of the circle = 2πr = 2 * 3.14 * 1.25 = 7.85m

⇒The time to describe this circle t = 2πr/ v

⇒ in 1 revolution the angle θ = 2π radians

This angle is subtended in time t = 2πr/ v

⇒ The angular speed = ω = θ/t = 2π ( v/ 2πr) = v/r = 2/1.25 = 1.6 rad/s

⇒ The angular momentum of man = I*ω = 62.5 * 1.6 = 100

Since the angular momentum is conserved, before and after the man starts running we have :

Angular momentum of disk = angular momentum of the man

⇒ with Angular momentum of disk = Idisk ωdisk

⇒ Idisk = MdiskR

⇒ with Angular momentum of disk = 100

or I(disk)*ω(disk) = 100

I(disk) = M(disk)*R ²/2 = 100*2*2 / 2 = 200

⇒ with M(disk) = the mass of the disk = 1.00 * 10² kg

⇒ with R = the radius of the disk = 2.00 m

200 ωdisk = 100

ωdisk = 100/200 = 0.5 rad/s

The resulting angular speed of the disk is 0.5 rad/s

(Since the angular speed is positive, the rotation is counterclockwise)