Two small beads of mass m are free to slide on a frictionless rod of mass M and

Question

Two small beads of mass m are free to slide on a frictionless rod of mass M and length L, as shown in Fig. 13-34. Initially the beads are held together at the rod center, and the rod is set spinning freely with initial angular speed 0 about a vertical axis coming out of the page in Fig. 13-34. The beads are released, and they slide to the ends of the rod and then off. Figure 13-34. Find the expressions for the angular speed of the rod at the following times. (Use the following as necessary: m, M, L, and 0.) (a) when the beads are halfway to the ends of the rod = (b) when they're at the ends = (c) after the beads are gone = Hint: Two of the answers are the same. Why?

Explanation / Answer

a)

Ir = moment of inertia of rod = ML2/12

Wo = initial angular speed

when the beads are at halfway :

Ibeads = moment of inertia of two beads = 2m(L/2)2 = (0.5) m L2

Wf = final angular speed

using conservation of angular momentum

Ir Wo = (Ir + Ibeads) Wf

(ML2/12) Wo = (ML2/12 + (0.5) m L2) Wf

Wf = (ML2/12) Wo / ((ML2/12 + (0.5) m L2))

b)

when at the end

Ibeads = moment of inertia of two beads = 2m(L)2 = 2 m L2

Wf = final angular speed

using conservation of angular momentum

Ir Wo = (Ir + Ibeads) Wf

(ML2/12) Wo = (ML2/12 + 2 m L2) Wf

Wf = (ML2/12) Wo / ((ML2/12 + 2 m L2))

c)

just after leaving the ends

Ibeads = moment of inertia of two beads = 2m(L)2 = 2 m L2

Wf = final angular speed

using conservation of angular momentum

Ir Wo = (Ir + Ibeads) Wf

(ML2/12) Wo = (ML2/12 + 2 m L2) Wf

Wf = (ML2/12) Wo / ((ML2/12 + 2 m L2))

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