physics angular momentum problem?

Question

two disks are rotating about the same axis. disk A has a moment of inertia of 3.4 kg m^2 and an angular velocity of +7.2 rad/s. disk B is rotating with an angular velocity of -9.8 rad/s. the two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.4 rad/s. the axis or rotation for this unit is the same as that for the seperate disks. what is the moment of inertia of disk B?

Answer 1

L=I*w
Conservation of angular momentum: Li=Lf
-->3.4*7.2+I*-9.8=(3.4+I)*-2.4
solve for I:
I = 4.4

Answer 2

purely finished this, so i assume. Madison via twist of destiny? B. L = r x p. be conscious the radius from the axis of rotation end of the rod is d/2. the preliminary momentum is the mass of the ball cases the preliminary velocity. L = mv(d/2). v preliminary this is C. I of a rod is (a million/12)ML^2. I of a ball is mr^2. L = d and r = d/2, so which you would be able to alter those in. The moments of inertia would be further so Itotal = (a million/12)Md^2 + m(d/2)^2 D. L = Iw. i grew to become into desperate in C so purely multiply via omega. E. Use your solutions from b and d to resolve for omega. F. KE = (a million/2)mv^2 G. KEfinal = (a million/2)Iw^2. utilising the entire 2d of inertia and omega from C and E and a few simplifying you will get (3(m^2)(v^2))/(2(M+3m)) H. (KEinitial - KEfinal)/KEinitial

Answer 3

You can use conservation of angular momentum
IA*wA+IB*wB=(IA+IB)*wf
wA=7.2
wB=-9.8
wf=-2.4
IA=3.4
so
3.4*7.2-IB*9.8=-(3.4+IB)*2.4
doing some algebra
3.4*(7.2-2.4)=IB(9.8-2.4)

j