1. ## Newtonian Mechanics

A uniform rod of mass m, is placed at right angles to a smooth plane of inclination αwith one end in contact with it. The rod is then released. Assume that the angle madeby the rod with the inclined plane at any time is equal to θ as shown in Figure Q7. Assume that the rod is released. Implement a suitable coordinate axis and writedown the equation of motion for the Centre of Gravity, G of the rod. Need help with the coordinate axis part. Im able to solve for the equation of motion for the centre of gravity,g of the rod using torque.

2. ## Re: Newtonian Mechanics

Perhaps they mean that the incline should be considered the "x"-axis and the perpendicular line to that incline the "y"-axis. This would require you to change the direction of gravity (as the $x$- and $y$-components of gravity would be different from normal).

3. ## Re: Newtonian Mechanics

hmm okay i'll try working with that. for the next part of the question, assuming that the radius of gyration of the rod about the centre of inertia k is
given by k^2= a^2/3, write down the equation of energy.
For the earlier part,the answer i gotten was m(a^2/3)theta doubledot= -Ra costheta, where R is the normal reaction force and i assumed the length of the rod to be 2a.

4. ## Re: Newtonian Mechanics

Originally Posted by noobpronoobpro
hmm okay i'll try working with that. for the next part of the question, assuming that the radius of gyration of the rod about the centre of inertia k is
given by k^2= a^2/3, write down the equation of energy.
For the earlier part,the answer i gotten was m(a^2/3)theta doubledot= -Ra costheta, where R is the normal reaction force and i assumed the length of the rod to be 2a.
Can you write this out in LaTeX? It is very difficult to read the way you are writing it. Look up LaTeX examples to see how to format it. It is very easy to learn, and it will make a world of difference in communicating questions on these forums.

5. ## Re: Newtonian Mechanics

Using LaTeX, you have written:

$m \frac {a^2} 3 \ddot \theta = -Ra \cos \theta$

Two problems with this:

1. On the left side 'a' is the length of the rod, but on the right side 'a' is half the length of the rod. You must be consistent.
2. This equation has to do with torque - but the contact force does not contribute to torque about the contact point. What you are missing is the force of gravity, which creates a torque about the contact point equal to the weight of the rod times the distance from the contact point to the center of gravity of the rod times the angle of the rod to the vertical (which is alpha plus theta).

6. ## Re: Newtonian Mechanics

The torque law states that:
http://latex.codecogs.com/png.latex?...a%20=%20torque
by using that with https://wikimedia.org/api/rest_v1/me...f4706aeae6747e with L=2a, i got the left side of the equation. but y doesn't the contact force contribute to torque about the contact point?

7. ## Re: Newtonian Mechanics

but y doesn't the contact force contribute to torque about the contact point?
... why doesn't a door swing open if one pushes on the hinges?

$\tau = r \times F$ ... $r=0$ at the pivot point.

8. ## Re: Newtonian Mechanics

This equation has to do with torque - but the contact force does not contribute to torque about the contact point. What you are missing is the force of gravity, which creates a torque about the contact point equal to the weight of the rod times the distance from the contact point to the center of gravity of the rod times the angle of the rod to the vertical (which is alpha plus theta).

Can you explain how do i get the last part, which is the angle of the rod to the vertical. By the vertical, are you referring to drawing a vertical line straight from the centre of gravity of the rod down to the inclined plane?

9. ## Re: Newtonian Mechanics

reference the attached diagram ...

Let $L$ = rod length and $m$ = rod mass

$\tau = mg \cdot \dfrac{L}{2} \cdot \sin{\color{red}{\phi}}$

note $\sin[90-(\theta-\alpha)] = \cos(\theta-\alpha)$ ...

$\tau = mg \cdot \dfrac{L}{2} \cdot \cos(\theta - \alpha)$

$\tau = I \cdot \dfrac{d^2{\theta}}{dt^2} = \dfrac{mL^2}{3} \cdot \dfrac{d^2{\theta}}{dt^2} = mg \cdot \dfrac{L}{2} \cdot \cos(\theta - \alpha)$

$\dfrac{d^2{\theta}}{dt^2} = \dfrac{3g}{2L} \cdot \cos(\theta - \alpha)$

note $\dfrac{d^2{\theta}}{dt^2}$ is a maximum when $\theta = \alpha$ as shown in the diagram you posted.

10. ## Re: Newtonian Mechanics

okay thanks for the explanation. greatly appreciated!