# A uniform slender rod of mass m and length L is released from rest, with its lower end touching a frictionless floor. At the initial moment, the rod is inclined at an angle theta = 30^0 with the vertical?

## Then the value of normal reaction from the floor just after the release will be? (please solve the question without using the concept of the instantaneous axis of rotation)

Mar 18, 2018

#### Answer:

$N = \frac{4}{7} m g$

#### Explanation:

If the floor is frictionless, the rod center of mass will move in a vertical line until the floor.

Calling

$m =$ rod mass
$l =$ rod length
$\theta =$ angle with the floor
$y = \frac{l}{2} \sin \theta$ mass center vertical coordinate.
${y}_{0} = \frac{l}{2} \sin {\theta}_{0}$
$g =$ gravity acceleration
$N =$ normal force
$J = \frac{1}{12} m {l}^{2}$ inertia rod moment regarding it's center of mass

The dynamics are

-m g + N = m ddot y = m l/2d^2/(dt^2)(sin theta) = l/2m (cos theta ddot theta - sin theta )dot theta^2

The energy conservation states

$m g {y}_{0} = m g y + \frac{1}{2} m {\dot{y}}^{2} + \frac{1}{2} J {\dot{\theta}}^{2}$

but

$\dot{y} = \frac{l}{2} \cos \theta \dot{\theta}$ then

$m \frac{l}{2} g \sin {\theta}_{0} = \frac{l}{2} m g \sin \theta + \left(\frac{1}{2} m {\left(\frac{l}{2} \cos \theta\right)}^{2} + \frac{1}{2} J\right) {\dot{\theta}}^{2}$ or

$m \frac{l}{2} g \sin {\theta}_{0} = \frac{l}{2} m g \sin \theta + \frac{1}{8} m {l}^{2} \left({\cos}^{2} \theta + \frac{1}{3}\right) {\dot{\theta}}^{2}$

dot theta^2 = (24 g (Sin theta_0 - Sin theta))/( l(5 + 3 Cos(2 theta)))

Deriving this expression we have

ddot theta =(6 g ( 3 (Cos(3 theta) + 4 Sin theta_0 Sin(2 theta))-19 Cos theta))/(l (5 + 3 Cos(2 theta))^2)

substituting into the dynamic equation we have finally

N =(2 g m (11 - 3 Cos(2 theta)- 12 Sin theta_0 Sin theta))/(5 + 3 Cos(2 theta))^2

Now for $\theta = {\theta}_{0} = \frac{\pi}{3}$

$N = \frac{4}{7} m g$