Equation of motion for a falling rod (with one end touching a frictionless surface) [closed]

Question

I have a quick question about the equation of motion for a falling rod (with one end touching a frictionless surface). The end touching the surface is not fixed. I am given the moment of inertia about the center of mass. Only gravity is acting on the rod. The rod begins at an angle $\theta$ above the ground.

I know that the rod will rotate about the center of mass, and the point touching the frictionless surface will slide along the surface, but I am having trouble calculating the torque. For reference, the left end of the rod is touching the surface, and the right end is in air. I have calculated the torque, $\tau$, from the right moment arm as $\tau=\frac{mg\cos{\theta}}{4I_G}$ because the right half of the rod contains half of the mass and half of the length. I don't really understand how to calculate the total net torque, though.

Any hints would be appreciated.

Answer 1

Since the surface is frictionless there is only vertical force. The torque is given by the normal force of the surface multiplied by the horizontal distance to the center of mass (c.o.m.). Now the normal force depends on the vertical acceleration of the c.o.m. - you know that the acceleration of the c.o.m. is a result of all the forces acting on the object, in this case just $F_n-mg$.

Now you just have to write the relationship between the two - torque gives rise to angular acceleration, which in turn results in changes in the vertical acceleration. For mass $m$, length $2\ell$, moment of inertia $I = \frac{1}{3}m\ell^2$ (rotation about the center of mass!), angle $\theta$ to the vertical (vertical: $\theta=0$), we can write the following equations:

Angular acceleration: $$I\ddot\theta = F_n \ell \sin\theta\\ \frac13m\ell^2 \ddot\theta= F_n\ell\sin\theta$$ $$\ddot\theta = \frac{3F_n\sin\theta}{m\ell}\tag1$$ Vertical acceleration of c.o.m.: $$y = \ell \cos\theta$$ $$\ddot{y} = -\ddot\theta\ell\cos\theta \tag2$$

But also we know $$\Gamma = I\ddot\theta\tag3$$ $$F_n - mg = m\ddot{y}\tag4$$

Eliminating $\ddot\theta$ from $(1)$ and $(2)$, and substituting the resulting expression for $\ddot{y}$ into $(4)$, we get

$$F_n = mg - 3F_n\sin\theta\cos\theta\\ =\frac{mg}{1+3\sin\theta\cos\theta}$$

And finally the torque follows:

$$\Gamma = F_n\ell\sin\theta\\ = \frac{mg\ell\sin\theta}{1+3\sin\theta\cos\theta}$$

Quick sanity check: when $\theta$ is close to $0$, there is little torque; the denominator would become zero when $3\sin\theta\cos\theta = -1$ - but that doesn't happen when $\theta\in[0,\pi/2]$ which is reassuring. In fact the plot of torque looks like this:

It's possible I made a mistake in the above, but it looks reasonable. The approach should be correct...

Answer 2

See https://physics.stackexchange.com/a/90894/392 for details on a very similar question.

If the body is in contact with the ground like this

then the equations of motion are

$$ \begin{aligned} F & = m \ddot{x}_C \\ N - m g & = m \ddot{y}_C \\ N \frac{\ell}{2} \sin\theta + F \frac{\ell}{2} \cos\theta & = I_C \ddot\theta \end{aligned} $$

with motion constraints

$$ \begin{aligned} \dot{x}_C & = \dot{x}_A - \frac{\ell}{2}\cos\theta \dot\theta & \ddot{x}_C & = \ddot{x}_A - \frac{\ell}{2}\cos\theta \ddot\theta + \frac{\ell}{2}\sin\theta \dot{\theta}^2 \\ \dot{y}_C & = - \frac{\ell}{2}\sin\theta \dot\theta & \ddot{y}_C & = - \frac{\ell}{2}\sin\theta \ddot\theta- \frac{\ell}{2}\cos\theta \dot{\theta}^2 \end{aligned} $$

and contact properties

$$ F = 0 \\ N > 0 $$

The above is solved by

$$ \boxed{ \begin{aligned} \ddot\theta & = \frac{ m \frac{l}{2} \sin\theta \left( g -\frac{l}{2} \dot{\theta}^2 \cos\theta \right)}{I_C + m \left(\frac{l}{2}\right)^2 \sin^2\theta} \\ N & = I_C \frac{ m \left( g -\frac{l}{2} \dot{\theta}^2 \cos\theta \right)}{I_C + m \left(\frac{l}{2}\right)^2 \sin^2\theta} \\ \ddot{x}_C & = 0 \\ \ddot{y}_C & = - \frac{ \frac{\ell}{2} \left( I_C \dot{\theta}^2 \cos\theta + m \frac{\ell}{2} g \sin^2\theta \right)}{I_C + m \left(\frac{l}{2}\right)^2 \sin^2\theta} \\ \end{aligned} } $$

Now the torque about the center of mass is

$$ \begin{aligned} \tau_C & = N \frac{\ell}{2} \sin\theta + F \frac{\ell}{2} \cos\theta \\ & = \frac{\ell}{2} N \sin\theta \\ & = I_C \frac{ m \frac{l}{2} \sin\theta \left( g -\frac{l}{2} \dot{\theta}^2 \cos\theta \right)}{I_C + m \left(\frac{l}{2}\right)^2 \sin^2\theta} \end{aligned} $$

^NOTE: The notation $\dot{x}_C$ and $\ddot{x}_C$ means the velocity and acceleration of point C along the x direction. Similarly for the rest of the velocity/acceleration components above. Notice how much more complex this problem is, than you might have originally thought.

Answer 3

I assume that the initial horizontal velocity is zero. Otherwise, we could just change the frame of reference.

If the surface is horizontal then this is effectively a constraint system with just one degree of freedom:

1. The constraint force has vertical direction.
2. The gravitational force has vertical direction.
3. No friction. This would be the only force in horizontal direction.

For these reasons you will only have acceleration of the center of mass in vertical direction and you can concentrate your investigations to the vertical motion.

Even if we do not use it below, some notes on the generation of torque: The constraint force acts on the contact point. The counter force is the inertial force that acts on the center of mass. The effective lever length is $\frac l2\cos(\theta)$.

Nevertheless, I would not like to write down balance equations. I would prefer some principle of constraint mechanics. For an instance Lagrange's principle.

You can use $\theta$ as generalized coordinate if you like.

The potential energy is: $$ U= mg\frac l2\sin(\theta) $$ the kinetic energy is $$T = \frac m2 \dot h^2 + \frac{J}2\dot\theta^2$$ with $h(\theta(t)) = \frac l2\sin(\theta(t))$, $\frac{d}{dt}h(\theta(t)) = \frac l2\cos(\theta)\dot\theta$ and with the moment of inertia $J$ for rotations about the center of mass. I do not elaborate that because it depends on the rod.

$$ T = \frac {ml}8 \cos^2(\theta)\dot\theta^2 + \frac{J}2\dot\theta^2 $$ The Lagrangian is $$ L(\theta,\dot\theta) = T-U = \frac {ml}8 \cos^2(\theta)\dot\theta^2 + \frac{J}2\dot\theta^2 - mg\frac l2\sin(\theta) $$ and Lagrange's equation is $$ \frac{d}{dt}\left(\partial_{\dot\theta} L\right)-\partial_\theta L = 0. $$ This is acually the equation of motion. $$ \frac{d}{dt}\left(\left(\frac{ml}{4}\cos^2(\theta)+J\right)\dot\theta\right) -\left(-\frac{ml}4\cos(\theta)\sin(\theta)\dot\theta^2-mg\frac l2\cos(\theta)\right)=0 $$ I hope that I did not make many mistakes and leave the test and the rest to you.