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Chapter 13 - Rotation II: A Conservation Approach

Moment of inertia for extended objects

So far our definition of moment of inertia is really only practical for systems comprised of one or more point like objects at a distance from an axis of rotation.

For extended objects we are much better served by considering an object as being made up of infinitesimally small mass elements, each a distance $R$ from the axis of rotation, and integrating over these mass elements to find the moment of inertia.

$I= \Sigma_{i} m_{i}R_{i}^{2}$ → $I=\int R^2\,dm$

Hoop, thin walled cylinder and solid cylinder

Considering rotation axis through the center of the circle

$I=\int R^2\,dm$

Hoop

$I=\frac{MR^2}{2\pi}\int_{0}^{2\pi}\,d\theta=MR^2$

Thin walled cylinder

$I=\frac{MR^2}{2\pi h}\int_{0}^{h}\int_{0}^{2\pi}\,d\theta\,dz=MR^2$

Solid Cylinder

$I=\frac{M}{\pi h R_{C}^2}\int_{0}^{R_{C}}\int_{0}^{h}\int_{0}^{2\pi}R^3\,d\theta\,dz\,dR=\frac{M}{\pi h R_{C}^2}\frac{ 2\pi h R_{C}^4}{4}=\frac{1}{2}MR_{C}^{2}$

Parallel axis theorem

Usually a rotation axis that passes through the center of mass of an object will be one of the easiest to find the moment of inertia for, because as we saw in the last lecture the center of mass usually reflects the symmetry of the object. More moments of inertia on wikipedia.

If we know the moment of inertia of an object around an axis that passes through it's center of mass there is a theorem that can help us find the moment of inertia around a different axis parallel to the axis through the COM.

If the axis of rotation is a distance $h$ from the axis through the COM then

$I=I_{COM}+Mh^{2}$

where M is the total mass of the object.

 As an example, a solid sphere stuck to a turning pole will have moment of inertia $I=\frac{2}{5}MR^{2}+MR^2=\frac{7}{5}MR^2$

When can we approximate a mass as a point?

Frequently we will want to approximate a mass at some distance from it's center of rotation as a point mass. i.e we would like to simply write

$I=mR_{1}^{2}$

The parallel axis theorem tells us that in fact (if the mass is spherical and solid)

$I=mR_{1}^{2}+\frac{2}{5}mR_{2}^{2}$

The fractional error introduced by the above approximation is

$\frac{\frac{2}{5}mR_{2}^{2}}{mR_{1}^{2}+\frac{2}{5}mR_{2}^{2}}=\frac{\frac{2}{5}}{\frac{R_{1}^{2}}{R_{2}^2}+\frac{2}{5}}$

If we want to be accurate to say 1% we only need $\frac{R_{1}}{R_{2}}\approx\sqrt{40}\approx 6$

Atwood machine with rotation

 We can apply our new knowledge about moment of inertia to our old friend the Atwood machine.If we take in to account the mass $M$ and radius $R$ of the pulley the tensions in the ropes on either side of the pulley need not be the same.

The sum of the torques on the pulley will be given by

$\Sigma \tau =(T_{2}-T_{1})R$

and as we saw that the moment of inertia for solid cylinder is $I=\frac{1}{2}MR^2$ we can find the angular acceleration of the pulley

$\large \alpha=\frac{\Sigma \tau}{I}=\frac{2(T_{2}-T_{1})}{MR}$

This can be related to the tangential acceleration of a point on the edge of the pulley by multiplying by R as $a=\alpha R$

$\large a=\frac{\Sigma \tau}{I}=\frac{2(T_{2}-T_{1})}{M}$

As we did before when we neglected rotation we should write Newton's Second Law for the two weights

$m_{2}a=m_{2}g-T_{2}$ → $T_{2}=m_{2}g-m_{2}a$
$m_{1}a=T_{1}-m_{1}g$ → $T_{1}=m_{1}g+m_{1}a$

$T_{2}-T_{1}=m_{2}g-m_{2}a-m_{1}g-m_{1}a$

$\frac{1}{2}Ma=m_{2}g-m_{2}a-m_{1}g-m_{1}a$

$\large a=g\frac{m_{2}-m_{1}}{\frac{1}{2}M+m_{1}+m_{2}}$

Rotational Kinetic Energy

Each piece of mass in a rotation problem that has velocity $v$ should have kinetic energy

$K=\frac{1}{2}mv^2$

In terms of the angular velocity this is

$K=\frac{1}{2}m\omega^2R^2$

and if we sum over all the masses

$K=\frac{1}{2}(\Sigma m_{i}R_{i}^2)\omega^{2}=\frac{1}{2}I\omega^2$

Conservation of energy with rotation

For a rolling object $v=\omega r$

The kinetic energy of a rolling object is therefore

$\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}I\frac{v^{2}}{r^2}$

The kinetic energy thus depends on the moment of inertia of the object.

Hoop and Disk

Suppose we release a hoop and disk from the top of a slope. They begin with the same potential energy, which one gets to the bottom of the slope first?

Hoop and disk solution

Hoop

$\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}mr^{2}\frac{v^{2}}{r^2}=mv^{2}$

$mgh=mv^{2}$

$v=\sqrt{gh}$

Disk

$\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}\frac{1}{2}mr^{2}\frac{v^{2}}{r^2}=\frac{3}{4}mv^{2}$

$mgh=\frac{3}{4}mv^{2}$

$v=\sqrt{\frac{4}{3}gh}$

Recall that for a sliding object (without friction) $v=\sqrt{2gh}$

Solid sphere $I=\frac{2}{5}mr^{2}$
Hollow sphere $I=\frac{2}{3}mr^{2}$

Work energy theorem for rotation

$W=\int \vec{F}\cdot\,d\vec{l}=\int F_{\perp}R\,d\theta=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta$

$\tau=I\alpha=I\frac{d\omega}{dt}=I\frac{d\omega}{d\theta}\frac{d\theta}{dt}=I\omega\frac{d\omega}{d\theta}$

$W=\int_{\omega_{1}}^{\omega_{2}}I\omega\,d\omega=\frac{1}{2}I\omega_{2}^2-\frac{1}{2}I\omega_{1}^2$

Therefore the work done in rotating an object through an angle $\theta_{2}-\theta_{1}$ is equal to the change in the rotational kinetic energy of the object.

Power and Torque

$W=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta$

$P=\frac{dW}{dt}=\tau\frac{d\theta}{dt}=\tau\omega$

This equation can help us understand the two “figures of merit” often given for a car engine, horsepower and torque.

Angular Momentum

Linear momentum

$\vec{p}=m\vec{v}$

By analogy we can expect angular momentum is given by

$L=I\omega$

Units $\mathrm{kgm^{2}/s}$

Newton's Second Law for translational motion

$\Sigma \vec{F} = m \vec{a}=\frac{d\vec{p}}{dt}$

By analogy we can expect Newton's Second Law for rotational motion is given by

$\Sigma \tau=I \alpha=\frac{dL}{dt}$

Conservation of angular momementum

In the absence of a net external torque

$\Sigma \tau=\frac{dL}{dt}=0$

and angular momentum is conserved.

$L=I\omega=\mathrm{constant}$

Changing the moment of inertia of a spinning object

Suppose I with two weights in my hands can be approximated by an 80 kg cylinder of radius 15 cm. My moment of inertia if I am spinning around an axis going down my center will be

$I=\frac{1}{2}MR^{2}=0.9\,\mathrm{kgm^2s^{-1}}$

With my arms (which we approximate as 3.5 kg and a length of 0.75 m from my shoulder) extended holding 2.3 kg weights my moment of inertia will be considerably higher. When I am holding my arms and weights out I should remove their mass from the cylinder

$I=\frac{1}{2}68.4\times0.15^2=0.77\,\mathrm{kgm^2s^{-1}}$

The moment of inertia of the two weights around the axis when my arms are extended is

$I=2\times2.3\times(0.9)^{2}=3.726\,\mathrm{kgm^2s^{-1}}$

The moment of inertia of an arm if it is rotated around it's center of mass is

$I=\frac{1}{12}\times3.5\times(0.75)^2=0.16\,\mathrm{kgm^2s^{-1}}$

But if it rotates around a point $0.375+0.15\,\mathrm{m}$ from it's center of mass then its moment of inertia is

$I=\frac{1}{2}\times3.5\times(0.75/2)^2+3.5\times(0.375+0.15)^2$
$=0.16+0.96=1.12\,\mathrm{kgm^2s^{-1}}$

So the total moment of inertia due to the extended arms, weights and the cylinder is

$I=0.77+3.726+2.24=6.736\,\mathrm{kgm^2s^{-1}}$

Consequence of conservation of angular momentum

From the previous calculation we have

With arms in $I=0.9\,\mathrm{kgm^2s^{-1}}$

With arms out $I=6.736\,\mathrm{kgm^2s^{-1}}$

If I am rotating with angular velocity $\omega$ with my arms out and I bring them in then my angular velocity afterward is $\omega'$ and from conservation of momentum

$L=I\omega=I'\omega'$

$\frac{\omega'}{\omega}=\frac{I}{I'}=\frac{6.916}{0.9}=7.5$

Let's see if it works!

Angular momentum as a vector cross product

The most general definition of angular momentum is as the cross product of the position vector $\vec{r}$ and the linear momentum of the object $\vec{p}$. Here we consider a single particle of mass $m$

$\vec{L}=\vec{r}\times\vec{p}$

We can relate the angular momentum to the torque by taking the derivative of $\vec{L}$ with respect to time

$\frac{d\vec{L}}{dt}=\frac{d}{dt}(\vec{r}\times\vec{p})=\frac{d\vec{r}}{dt}\times\vec{p}+\vec{r}\times\frac{d\vec{p}}{dt}=\vec{v}\times m\vec{v}+\vec{r}\times\frac{d\vec{p}}{dt}=\vec{r}\times\frac{d\vec{p}}{dt}$

In an inertial reference frame (non accelerating reference frame) $\Sigma \vec{F}=\frac{d\vec{p}}{dt}$

$\vec{r}\times\frac{d\vec{p}}{dt}=\vec{r}\times\Sigma\vec{F}=\Sigma\vec{\tau}$

$\Sigma\vec{\tau}=\frac{d\vec{L}}{dt}$

Angular momentum of a system of objects

For a system of objects the total angular momentum is given

$\frac{d\vec{L}}{dt}=\Sigma{\vec{\tau}_{ext}}$

following from the usual cancellation of internal forces between objects due to Newton's Third Law.

This equation, like the one before, is only true when $\vec{\tau}_{ext}$ and $\vec{L}$ are calculated about a point which is moving uniformly in an inertial reference frame.

If these quantities are calculated around a point that is accelerating the equation does not hold, except for in one special case, which is for motion around the center of mass of the system (proof in text). So we can say that

$\frac{d\vec{L}_{CM}}{dt}=\Sigma\vec{\tau}_{CM}$

even if the center of mass is accelerating. This is very important!

Angular momentum of rigid objects

 If we have a rigid object rotating about an axis that has a fixed direction it is useful to know the component of angular momentum along the axis of the rotating object. Each piece of the object will have angular momentum $\vec{L}_{i}=\vec{r}_{i}\times\vec{p}_{i}$

We can express the component along the rotation axis of each of these individual angular momenta as

$L_{i\omega}=r_{i}p_{i}\cos\phi=m_{i}v_{i}r_{i}\cos\phi$

Using $r_{i}\cos\phi=R_{i}$ and $v_{i}=R_{i}\omega_{i}$ we find that

$L_{i\omega}=m_{i}R_{i}^{2}\omega$

Summing over the entire object

$L_{\omega}=(\Sigma_{i}m_{i}R_{i}^2)\omega=I\omega$

This looks like the intuitive relationship we used earlier, but we should be careful, this equation is for the component of angular momentum along the rotation axis. If the object is symmetric we can however reason that all components of angular momentum not along the axis cancel out and

$\vec{L}=I\vec{\omega}$

If the conditions discussed above are fulfilled we can also use $\Sigma\tau=\frac{dL}{dt}$ to show that

$\Sigma\tau_{axis}=\frac{d}{dt}(I\omega)=I\frac{d\omega}{dt}=I\alpha$

Some fun with a bike wheel

If I want to turn the bike wheel from spinning vertically to horizontally which way should I jerk the handle of the wheel.

A. Up B. Down C. To my left d. To my right

The gyroscope

As we saw with a spinning bike wheel it takes a large torque to reorient a large angular momentum vector. If a spinning object is mounted so that it is free to orient itself in any direction, as it is in a gyroscope then the direction of it's angular momentum is constant and this can be used a useful reference, for example in an airplane or boat.

In a gyroscope the freedom for the spinning wheel to orient is achieved by its suspension in pivoted supports called gimbals.

Gyroscope Precession

 If we apply a torque to a gyroscope by hanging a weight from it's axis an initially surprising effect is observed. Rather than falling we see the gyroscope begins to undergo precession, i.e the direction of the axis of rotation begins to change. This result can be understood by looking at the direction of the torque that is generated.

$\vec{\tau}=\vec{r}\times m\vec{g}$

By inspection we can see that independent of the orientation of the gyroscopes axis the torque will always be directed parallel to the surface of the Earth and perpendicular to the initial angular momentum vector. This torque acts to change the direction, but not the magnitude of the angular momentum vector. We'll consider this problem in the mathematically easiest situation which is when the axis of rotation is horizontal (a more general derivation is in your textbook). We define an angle $\theta$ to describe the orientation of the angular momentum vector in the horizontal plane, and then express the

$dL=L\,d\theta$

If we define the angular velocity of the precession as $\Omega=\frac{d\theta}{dt}$ then

$\Omega=\frac{1}{L}\frac{dL}{dT}=\frac{\tau}{L}=\frac{mgr}{L}$

Gyroscope Nutation

If the precession above seems just a little too “magical” to you then you have good instincts! In actual fact to be able to precess the axis of the gyroscope actually has to drop a little, as there is an angular momentum associated with the precession. Without this drop it is impossible for the gyroscope to precess! Indeed, the axis of the gyroscope will actually drop beyond the equilibirum position and will be pulled back up, again overshooting and executing an oscillatory motion around the mean position of the plane of precession that we call nutation.

When the angular momentum of the gyroscope is high this drop is barely noticeable, because the angular velocity of precession, which is inversely proportional to $L$ ($\Omega=\frac{mgr}{L}$), is quite small. Nutation at high angular velocities is therefore a barely noticeable shaking of the axis of rotation. At lower angular momentum the nutation is much more obvious.

In the absence of friction the path that the tip of angular momentum vector would draw out is a cycloid (upside down), though as the gimbal bearings have quite a bit of friction the nutation oscillations are relatively quick to damp down leaving the smooth precession we discussed initially.