A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane. There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical…

Question

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane. There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is

Accepted Answer

(1) (3g)/(2l)sin θ. Take moments about the pivot, so the pivot reaction drops out. The weight Mg acts at the centre, a distance l/2 along the rod. With θ measured from the vertical, the perpendicular distance from the pivot to the line of the weight is l/2 sin θ. τ=Mg·l/2 sin θ. Rod about one end: I=(Ml²)/3. α=τ/I=(Mgl/2 sin θ)/((Ml²)/3)=(3g)/(2l)sin θ. Sanity check: at θ=0 the rod is upright and α=0, an unstable balance; at θ=π/2 the rod is horizontal and α is at its largest, (3g)/(2l).

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