Since this blog is meant to concentrate on these not emphasized in textbooks or offer a slightly different way of looking at old things, we now come to place where I think a great deal of material has been written on the topic and most of it quite good. Now, we derive Euler's rigid body equations in all their glory, I found a few interesting problems, which relate to rotational motion.
1) You are to imagine a rod pivoted to a horizontal beam but free to swing in a plane perpendicular to the plane containing the beam. The rod has zero width, and a density not necessarily constant. Its moment of inertia about an axis parallel to the beam is I and the center o f mass is located distance b from the pivot. Find the equation of motion for small swings, its period and the length of a simple pendulum that would have the same period.
This is called a compound pendulum( contrast it with the simple pendulum where all the mass is concentrated at a point). Let \(\Gamma\) be the torque and L is the angular momentum.
\begin{align}
\dot{L}&= \Gamma \\
I \dot{\omega}& = - bmg\sin \theta \\
\ddot{\theta} &= -bg \theta m \\
\ddot{\theta} &= \frac{-bg\theta m}{I}\\
\end{align}
It should b:e obvious that the \(\omega\) or angular frequency is\( \sqrt{\frac{mbg}{I}}\) with period \(2 \pi\sqrt{\frac{I}{mbg}}\). Now the angular frequency of a simple pendulum is \( \sqrt{\frac{g}{l}}\). Setting the expressions for the different angular frequencies equal gives that the length of the simple pendulum must be \(\frac{I}{mb}\).
2) Imagine the same rod as above but this time it is hit with a horizontal force a distance c from the pivot with impulse F\(\Delta \)T = \(\xi\). What is the angular momentum delivered to the pendulum, pivot just after it has been hit? What is special distance \(b_o\) such that the impulse delivered to the pivot is zero?
Mathematically, I do not think this is a challenging problem. It is the conceptual physics involved that makes it quite challenging. So here is my solution:
We can treat the pendulum as though all its mass is centered at the center of mass. Now since it is not moving before we hit it, the initial velocity is zero. But this implies that the impulse is equal to the moment just after we hit it.(Just consider newton's second law for discrete times). Thus the angular momentum after it has been hit is simply \(b \xi\). Now the impulse delivered to the pivot will be m\(v_p\) where \(v_p\) is the velocity at the pivot. The velocity at the pivot is the velocity of the center of mass \( v_{cm} - v_{h}\) where \(v_h\) is the tangential velocity at the point it is hit. Therefore impulse is : \( mv_{cm} -m c\omega\) = \( \xi - \frac{mbc\xi}{I}\). Remember L = \(I\omega\), c is the distance from the pivot to where it is hit and b is the distance from pivot to center of mass. If we want to find that special distance \(b_o\) where impulse at the pivot is zero just set the expression for the impulse equal to zero and you will get that \( b_o = \frac{I}{mb}\)
1) You are to imagine a rod pivoted to a horizontal beam but free to swing in a plane perpendicular to the plane containing the beam. The rod has zero width, and a density not necessarily constant. Its moment of inertia about an axis parallel to the beam is I and the center o f mass is located distance b from the pivot. Find the equation of motion for small swings, its period and the length of a simple pendulum that would have the same period.
This is called a compound pendulum( contrast it with the simple pendulum where all the mass is concentrated at a point). Let \(\Gamma\) be the torque and L is the angular momentum.
\begin{align}
\dot{L}&= \Gamma \\
I \dot{\omega}& = - bmg\sin \theta \\
\ddot{\theta} &= -bg \theta m \\
\ddot{\theta} &= \frac{-bg\theta m}{I}\\
\end{align}
It should b:e obvious that the \(\omega\) or angular frequency is\( \sqrt{\frac{mbg}{I}}\) with period \(2 \pi\sqrt{\frac{I}{mbg}}\). Now the angular frequency of a simple pendulum is \( \sqrt{\frac{g}{l}}\). Setting the expressions for the different angular frequencies equal gives that the length of the simple pendulum must be \(\frac{I}{mb}\).
2) Imagine the same rod as above but this time it is hit with a horizontal force a distance c from the pivot with impulse F\(\Delta \)T = \(\xi\). What is the angular momentum delivered to the pendulum, pivot just after it has been hit? What is special distance \(b_o\) such that the impulse delivered to the pivot is zero?
Mathematically, I do not think this is a challenging problem. It is the conceptual physics involved that makes it quite challenging. So here is my solution:
We can treat the pendulum as though all its mass is centered at the center of mass. Now since it is not moving before we hit it, the initial velocity is zero. But this implies that the impulse is equal to the moment just after we hit it.(Just consider newton's second law for discrete times). Thus the angular momentum after it has been hit is simply \(b \xi\). Now the impulse delivered to the pivot will be m\(v_p\) where \(v_p\) is the velocity at the pivot. The velocity at the pivot is the velocity of the center of mass \( v_{cm} - v_{h}\) where \(v_h\) is the tangential velocity at the point it is hit. Therefore impulse is : \( mv_{cm} -m c\omega\) = \( \xi - \frac{mbc\xi}{I}\). Remember L = \(I\omega\), c is the distance from the pivot to where it is hit and b is the distance from pivot to center of mass. If we want to find that special distance \(b_o\) where impulse at the pivot is zero just set the expression for the impulse equal to zero and you will get that \( b_o = \frac{I}{mb}\)
No comments:
Post a Comment