Question

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 1

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 :-)

Answer 2

First, by the conservation of Torque,

we must get the angular acceleration 'α' in terms of F, L and m; which is depicted in the picture below.

we get angular acceleration α = 12F/7mL

(I got the moment of inertia at the point of application of force by using parallel axis theorem)

Now,

By applyong laws of motion in rotation,

θ = ω₀t + ¹/₂ α t² (similar to the formula in linear motion; s = ut + ¹/₂ at²)

where θ → angular displacement (angle rotated in time 't')

ω₀ → initial angular velocity = 0

α → angular acceleration.

t → time

θ = ¹/₂ α t²

θ = ¹/₂ (12F/7mL) t²

∴ θ = 6Ft²/7mL

Thus, the angle rotated by the rod in time 't' is 6Ft²/7mL

Hope it helps!!