Question

A rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then find the maximum angle '$\theta$' in degrees made by the hinged upper half with the vertical.

Answer 1

Applying conservation of energy principle when rod is vertical ,
loss in potential energy = gain in kinetic energy + losses
Lets neglect the friction at hinge so losses are zero.
$\frac{mgL}{2}=\frac{1}{2}I{\omega }^2$
$I=\frac{m{L}^2}{3}$ (moment of inertia of rod about its upper end)
$\frac{mgL}{2}=\frac{1}{2}\left(\frac{1}{3}m{L}^2\right){\omega }^2$
$\frac{1}{3}m{L}^2{\omega }^2=mgL$
At vertical position when rod breaks at its middle point $L^{\prime }=\frac{L}{2}$
Again applying energy conservation principle between when rod reaches its maximum angle with vertical ($\theta$) and at its vertical position.
$\frac{I^{\prime }{\omega }^2}{2}=\frac{mgL}{4}(1-cos\theta )$
$I^{\prime }=\frac{m{L}^{\prime 2}}{3}=\frac{m{L}^2}{12}$
$\frac{m{L}^2{\omega }^2}{24}=\frac{mgL}{4}(1-cos\theta )$
$\frac{m{L}^2{\omega }^2}{3}=\frac{mgL}{4}(1-cos\theta )$
Bring the result of application of conservation of energy principle in first stage
$\frac{1}{3}m{L}^2{\omega }^2=mgL$
$\frac{1}{8}mgL=\frac{1}{4}mgL(1-cos\theta )$
$1-cos\theta =\frac{1}{2}$
$cos\theta =\frac{1}{2}$
$\theta ={60}^0$