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At t=0 a bead of mass m is placed at a distance \frac{L}{2} from the axis of rotation. Initial speed of the bead is zero with respect to the rod. The rod is rotated with a constant angular velocity o as shown in diagram. Find the velocity of bead w.r.t ground, when it leaves out of the rod. (Given L is the length of rod.)
Asked by aksharashere3 | 26 Jan, 2024, 07:04: PM
answered-by-expert Expert Answer

Figure shows the bead initially at a distance L/2 from left end of rod and rod ratates about an axis passing through left end .

Let us assume axis is perpendicular to plane of page.

Due to circular motion, bead experiences centrifugal force and moves towards right edge of rod.

when the bead is at a distance r from left end , centrifugal acceleration is

img

Above expression is written as

img

img

By integrating above expression from r = L/2 to r = L with initial condition v = 0 at x = L/2 ,

we get velocity of bead when it reaches right edge is

img

v = ω L

At the time of coming out of right edge , bead also has tangential velocity (ω L)

Hence, magnitude of  the resultant velocity is  √2 (ω L) as shown in figure

 

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