1.

A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its center. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the center, The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts?

Answer»

Given in the question :-


Rod MASS = m

Rod length = l

force acting perpendicular from the DISTANCE L/4



Now the torque which acts on the rod

T = F * \frac{L}{4}



Moment of INERTIA of the rod

I= \frac{mL^2}{12}



Angular ACCELERATION of the rod,

α = T/I

α = FL*12/4mL²

\alpha = \frac{3F}{mL}



At time t , angle rotation in the rod

\theta = 0 * t+\frac{1}{2} at^2

=1/2 × (3F/mL) × t²

θ= 3Ft²/2mL



Hope it Helps :-)


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