LenDiscos.problem Points: 1 A uniform disk A of mass mA= 7.6 kg turns at ,-+49 r

Question

LenDiscos.problem Points: 1 A uniform disk A of mass mA= 7.6 kg turns at ,-+49 rad/s about a fixed central axis. Another rotating disk B of mass mB= 10.3 kg, with the same radius R of disk A, is dropped onto the freely spinning disk A (see figure). They become coupled and turn together with their centers superposed, as shown in the figure, with an angular velocity '=+33 rads. (The moment of inertia of the disk is ld [ 1/2lmR2, where m is the mass, and R is the radius) . The angular velocity of disk B before the impact is: COA

Explanation / Answer

given :

mass of A=mA=7.6 kg

mass of B=mB=10.3 kg

radius of both the disks=R

angular velocity of A=wA=49 rad/s

let angular velocity of B=wB

final angular speed=w’=33 rad/s

moment of inertia of A=0.5*mass of A*radius of A^2

=0.5*mA*R^2

moment of inertia of B=0.5*mB*R^2

total moment of inertia of both the masses =0.5*mA*R^2+0.5*mB*R^2=0.5*(mA+mB)*R^2

as there is no external force or torque, total angular momentum is conserved.

angular momentum=moment of inertia*angular velocity

using conservation of angular momentum principle:

0.5*mA*R^2*wA+0.5*mB*R^2*wB=0.5*(mA+mB)*R^2*w’

==>mA*wA+mB*wB=(mA+mB)*w’

==>wB=((mA+mB)*w’-mA*wA)/mB

=((7.6+10.3)*33-7.6*49)/10.3

=21.194 rad/s

angular velocity of disk B before the impact is 21.194 rad/s

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