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Applications of angular momentum

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Kinetic energy and rolling motion

The kinetic energy \(T\) in a rotating rigid body of mass \(M\) and moment of inertia \(I\) can easily be determined by establishing the kinetic energy of its constituent parts:

(85)\[\begin{align} T_{\rm rot} = \sum_j \frac12 m_j v_j^2 = \frac12 \left( \sum_j m_j r_j^2 \right) \omega^2\ = \frac12 I \omega^2\, , \end{align}\]

where we again used that \(\omega\) is the same for all elements in a rigid body. This kinetic energy is purely due to the rotation around an axis. In addition, we can set the rigid body in linear motion, where the centre of mass has velocity \(\v\). This leads to a translational kinetic energy

(86)\[\begin{align} T_{\rm trans} = \frac12 M v^2\, . \end{align}\]

The total kinetic energy is the sum of these:

(87)\[ T = T_{\rm trans} + T_{\rm rot} = \frac12 M v^2 + \frac12 I\omega^2\, .\]

This becomes important when we want to analyse rolling motion, for example when a cylinder rolls down a ramp. Previously, we considered frictionless sliding blocks that convert their potential energy entirely into kinetic energy: \(mgh = \frac12 m v^2\), so the final velocity of the block is \(v = \sqrt{2gh}\). However, now we can consider the effect of the energy going into the rotational motion of the cylinder. We assume the no-slipping condition, which says that the friction between the ramp and the cylinder is large enough so that the cylinder rolls and does not slide. This gives us a relation between the angular speed of the rotation and the translational speed:

(88)\[\begin{align} v = R\omega\, , \end{align}\]

where \(R\) is the radius of the cylinder. Substituting this into equation (87) and equating \(T=Mgh\), we obtain

(89)\[\begin{align} Mgh = \frac12 M v^2 + \frac12 \frac{I}{R^2} v^2 = \frac12 \left(M + \frac{I}{R^2} \right) v^2\, . \end{align}\]

Solving for \(v\), we find

(90)\[\begin{align} v = \sqrt{\frac{2Mgh}{M + I/R^2}}\, . \end{align}\]

This is smaller than the sliding block case, and reduces to it when \(I\to0\). Note that this is true for cylinders with any mass distribution, not just the uniformly solid cylinder and the open tube. We can even consider spherical objects or egg-shaped rigid bodies. As long as it can obey the no-slipping condition this formula holds. For example, you can do rolling experiments with raw and hard-boiled eggs. Would you expect a difference in the final centre of mass velocity?

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Here is a useful thing to remember. For every quantity pertaining to linear motion, there is a corresponding quantity for rotational motion. This is captured in the table below:

	linear	circular
position	\(\r\)	\(\vec{\theta}\)
velocity	\(\v = \dot{\r}\)	\(\vec{\omega} = \dot{\vec{\theta}}\)
acceleration	\(\a = \dot{\v}= \ddot{\r}\qquad\)	\(\vec{\alpha} = \dot{\vec{\omega}} = \ddot{\vec{\theta}}\)
mass	\(m\)	\(I = mr^2\)
momentum	\(\p\)	\(\L = \r\times\p\)
force	\(\F = \dot{\p}\)	\(\vec{\tau}=\r\times\F = \dot{\L}\)
kinetic energy	\(\frac12 mv^2\)	\(\frac12 I \omega^2\)

Scattering problems

Let’s consider the role of angular momentum in the energy of a planet orbiting a star. The total energy for a planet with mass \(m\) in the gravitational potential of a star with mass \(M\) can be written as

(91)\[\begin{align} E = \frac12 m v^2 + V(r)\, . \end{align}\]

We will keep \(V(r)\) for now to keep the notation simple and clear. We will also use polar coordinates, \(r\) and \(\theta\), in which \(v^2\) becomes

(92)\[\begin{align} v^2 = v_r^2 + v_\theta^2 = \left(\frac{\d r}{\d t}\right)^2 + r^2\left(\frac{\d \theta}{\d t}\right)^2 \, . \end{align}\]

From equation (64) we deduce that

(93)\[\begin{align} r^2\left(\frac{\d \theta}{\d t}\right)^2 = \frac{L^2}{(mr)^2} \, , \end{align}\]

where \(L\) is the angular momentum of the planet. The direction of \(\L\) is perpendicular to the plane of the orbital motion. Substituting \(v^2\) into the kinetic energy of the planet, we obtain

(94)\[E = \frac12 m \left(\frac{\d r}{\d t}\right)^2 + \frac{L^2}{2mr^2} + V(r)\, .\]

It is worth pausing at this important expression. The angular momentum \(L\) is conserved for any central potential \(V(r)\), so \(L\) is just another constant in this expression. We managed to eliminate \(\theta\) from \(E\) altogether, which means that there is a one-to-one relation between \(r\) and \(\theta\). If we know the coordinate \(r\) of the planet, we should be able to work out the coordinate \(\theta\). Once we know \(r(t)\) we can deduce \(\theta(t)\), and from that we can find \(v_r\) and \(v_\theta\) by differentiation. In addition, for a central potential we know that the resulting force will always be in the radial direction (we assume that the origin is chosen at the centre of mass). Hence, the acceleration of the planet will be in the direction of the centre of mass. Therefore, we have reduced the problem of planetary motion to a one-dimensional problem for \(r\). It now makes sense to define a new effective potential that includes the angular momentum term:

(95)\[ E = \frac12 m \left(\frac{\d r}{\d t}\right)^2 + V_{\rm eff}(r) \qquad\text{with}\qquad V_{\rm eff}(r) = V(r) + \frac{L^2}{2mr^2} \, .\]

The angular momentum term provides a repulsive potential, since it is a positive quantity, and is sometimes called the centrifugal potential energy. It is purely a result of describing the problem in polar coordinates instead of cartesian coordinates. This is very similar to the fictitious centrifugal force, which appeared when we described circular motion in polar coordinates. In both cases this centrifugal component is an expression of the tendency of massive bodies to move in a straight line.

When we look at the case of the gravitational potential, we see that the centrifugal potential drops off faster, i.e., with \(1/r^2\), than \(V(r)\), which drops off with \(1/r\). That means that for large distances \(V(r)\) is the dominant potential, but very close to the central mass the centrifugal force will dominate.