A uniform rod of length L and a mass M is free to rotate on a frictionless pin p

Question

A uniform rod of length L and a mass M is free to rotate on a frictionless pin passing through one end, as depicted in the figure ( on the unclear part it says L/2). The rod is released from rest in the horizontal position.

(a) Derive an expression for the rod's angular speed when it reaches its lowest position.

(b) Derive an expression for the (i) tangential speed of the center of mass of the rod, and (ii) tangential speed of the lowest point on the rod when it is in the vertical position.

NOTE: I= 1/3 ML2 (2= a square exponent) for a rod with center of rotation at one end.

* Please show all work and calculations to receive credit for this problem*

Explanation / Answer

Use the conservation of energy for this problem. Let the lowest position of centre of mass of rod (say O) has zero potential energy. when the rod is horizontal, it has potential energy MgL/2 relative to O.

In vertical position the rod has rotational KE equal to I^2/2 where I is the moment of inertia about the pivot.

Equating these two energies:

                              MgL/2 = I^2/2

          but               I = (ML^2)/3

using this we get,     = (3g/L)                    Ans.

b)   Now consider the relation b/w linear and angular speeds:

      i)        v (CM) = r = L/2 = (3gL)/2         Ans.

      ii)        now the lowest point of rod has twice radius as compared to the CM.Thus it has a speed twice of the speed of CM (since = v/r is same for every point on rod)

Thus               2v (CM) = (3gL)                 Ans.

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