Chapter 11

a. ${\mu }_{\text{s}}\ge \frac{\text{tan}\theta }{1+(m{r}^2\text{/}{I}_{\text{CM}})}$; inserting the angle and noting that for a hollow cylinder $I_{\text{CM}}=m{r}^2,$ we have ${\mu }_{\text{s}}\ge \frac{\text{tan}60\text{°}}{1+(m{r}^2\text{/}m{r}^2)}=\frac{1}{2}\text{tan}60\text{°}=0.87;$ we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, so the condition isn’t satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition ${\mu }_{\text{s}}\ge \frac{1}{3}\text{tan}\theta =\frac{1}{3}\text{tan}60\text{°}=0.58.$ The value of 0.6 for ${\mu }_{\text{S}}$ satisfies this condition, so the solid cylinder will not slip.

From the figure, we see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have
$\overrightarrow{l}=\overrightarrow{r}\times \overrightarrow{p}=(0.4\text{m}\widehat{i}\text{)}\times (1.67\times {10}^{\mathrm{-27}}\text{kg}(4.0\times {10}^6\text{m}\text{/}\text{s})\widehat{j})=2.7\times {10}^{\mathrm{-21}}\text{kg}·{\text{m}}^2\text{/}\text{s}\widehat{k}$

$I_{\text{sphere}}=\frac{2}{5}m{r}^2,I_{\text{cylinder}}=\frac{1}{2}m{r}^2$; Taking the ratio of the angular momenta, we have:
$\frac{L_{\text{cylinder}}}{L_{\text{sphere}}}=\frac{I_{\text{cylinder}}{\omega }_0}{I_{\text{sphere}}{\omega }_0}=\frac{\frac{1}{2}m{r}^2}{\frac{2}{5}m{r}^2}=\frac{5}{4}$. Thus, the cylinder has $25\%$ more angular momentum. This is because the cylinder has more mass distributed farther from the axis of rotation.

Using conservation of angular momentum, we have
$I(4.0\text{rev/min})=1.25I{\omega }_{\text{f}},{\omega }_{\text{f}}=\frac{1.0}{1.25}(4.0\text{rev}\text{/}\text{min})=3.2\text{rev}\text{/}\text{min}$

The Moon’s gravity is 1/6 that of Earth’s. By examining Equation 11.12, we see that the top’s precession frequency is linearly proportional to the acceleration of gravity. All other quantities, mass, moment of inertia, and spin rate are the same on the Moon. Thus, the precession frequency on the Moon is
${\omega }_P\text{(Moon)}=\frac{1}{6}{\omega }_P\text{(Earth)}=\frac{1}{6}(5.0\text{rad}\text{/}\text{s})=0.83\text{rad}\text{/}\text{s}.$

No, the static friction force is zero.

The wheel is more likely to slip on a steep incline since the coefficient of static friction must increase with the angle to keep rolling motion without slipping.

The cylinder reaches a greater height. By Equation 11.4, its acceleration in the direction down the incline would be less.

All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.

The particle must be moving on a straight line that passes through the chosen origin.

Without the small propeller, the body of the helicopter would rotate in the opposite sense to the large propeller in order to conserve angular momentum. The small propeller exerts a thrust at a distance R from the center of mass of the aircraft to prevent this from happening.

The angular velocity increases because the moment of inertia is decreasing.

More mass is concentrated near the rotational axis, which decreases the moment of inertia causing the star to increase its angular velocity.

A torque is needed in the direction perpendicular to the angular momentum vector in order to change its direction. These forces on the space vehicle are external to the container in which the gyroscope is mounted and do not impart torques to the gyroscope’s rotating disk.

$v_{\text{CM}}=R\omega \Rightarrow \omega =66.7\text{rad/s}$

$\alpha =3.3\text{rad}\text{/}{\text{s}}^2$

$I_{\text{CM}}=\frac{2}{5}m{r}^2,a_{\text{CM}}=3.5\text{m}\text{/}{\text{s}}^2;x=15.75\text{m}$

positive is down the incline plane;
$a_{\text{CM}}=\frac{mg\text{sin}\theta }{m+(I_{\text{CM}}\text{/}{r}^2)}\Rightarrow I_{\text{CM}}=r^2\left[\frac{mg\text{sin}30}{a_{\text{CM}}}-m\right]$,
$x-x_0=v_{0}t-\frac{1}{2}{a}_{\text{CM}}{t}^2\Rightarrow a_{\text{CM}}=2.96{\text{m/s}}^2,$
$I_{\text{CM}}=0.66m{r}^2$

$\alpha =67.9\text{rad}\text{/}{\text{s}}^2$,
${(a_{\text{CM}})}_x=1.5\text{m}\text{/}{\text{s}}^2$

$W=\mathrm{-1080.0}\text{J}$

Mechanical energy at the bottom equals mechanical energy at the top;
$\frac{1}{2}m{v}_0^2+\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\left(\frac{v_0}{r}\right)}^2=mgh\Rightarrow h=\frac{1}{g}\left(\frac{1}{2}+\frac{1}{4}\right)v_0^2$,
$h=7.7\text{m,}$ so the distance up the incline is $22.5\text{m}$.

Use energy conservation
$\frac{1}{2}m{v}_0^2+\frac{1}{2}{I}_{\text{Cyl}}{\omega }_0^2=mg{h}_{\text{Cyl}}$,
$\frac{1}{2}m{v}_0^2+\frac{1}{2}{I}_{\text{Sph}}{\omega }_0^2=mg{h}_{\text{Sph}}$.
Subtracting the two equations, eliminating the initial translational energy, we have
$h_{\mathrm{Cyl}}=\frac{v_0^2}{g}$,
$h_{\mathrm{Sph}}=\frac{5v_0^2}{6g}$,
The ratio of the height reached by the sphere to the height reached by the cylinder is $\frac{h_{\text{Sph}}}{h_{\text{Cyl}}}=\frac{5}{6}.$
So the sphere reaches a lower height of $\frac{5}{6}\left(1.0\text{m}\right)=0.83\text{m.}$

The magnitude of the cross product of the radius to the bird and its momentum vector yields $rp\text{sin}\theta$, which gives $r\text{sin}\theta$ as the altitude of the bird h. The direction of the angular momentum is perpendicular to the radius and momentum vectors, which we choose arbitrarily as $\widehat{k}$, which is in the plane of the ground:
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}=hmv\widehat{k}=(300.0\text{m})(2.0\text{kg})(20.0\text{m}\text{/}\text{s})\widehat{k}=\mathrm{12,000.0}\text{kg}·{\text{m}}^2\text{/}\text{s}\widehat{k}$

a. $\overrightarrow{l}=45.0\text{kg}·{\text{m}}^2\text{/}\text{s}\widehat{k}$;
b. $\overrightarrow{\tau }=10.0\text{N}·\text{m}\widehat{k}$

a. ${\overrightarrow{l}}_1=\mathrm{-0.4}\text{kg}·{\text{m}}^2\text{/}\text{s}\widehat{k}$, ${\overrightarrow{l}}_2={\overrightarrow{l}}_4=0$,
${\overrightarrow{l}}_3=1.35\text{kg}·{\text{m}}^2\text{/}\text{s}\widehat{k}$; b. $\overrightarrow{L}=0.95\text{kg}·{\text{m}}^2\text{/}\text{s}\widehat{k}$

a. $L=1.0\times {10}^{11}\text{kg}·{\text{m}}^2\text{/}\text{s}$; b. No, the angular momentum stays the same since the cross-product involves only the perpendicular distance from the plane to the ground no matter where it is along its path.

a. $\overrightarrow{v}=\text{−}gt\widehat{j},{\overrightarrow{r}}_{\perp }=\text{−}d\widehat{i},\overrightarrow{l}=mdgt\widehat{k}$;
b. $\overrightarrow{F}=\text{−}mg\widehat{j},{\displaystyle \sum \overrightarrow{\tau }}=dmg\widehat{k}$; c. yes

a. $mgh=\frac{1}{2}m{(r\omega )}^2+\frac{1}{2}\frac{2}{5}m{r}^2{\omega }^2$;
$\omega =51.2\text{rad}\text{/}\text{s}$;
$L=16.4\text{kg}·{\text{m}}^2\text{/}\text{s}$;
b. $\omega =72.5\text{rad}\text{/}\text{s}$;
$L=23.2\text{kg}·{\text{m}}^2\text{/}\text{s}$

a. $I=720.0\text{kg}·{\text{m}}^2$; $\alpha =4.20\text{rad}\text{/}{\text{s}}^2$;
$\omega (10\text{s})=42.0\text{rad}\text{/}\text{s}$; $L=3.02\times {10}^4\text{kg}·{\text{m}}^2\text{/}\text{s}$;
$\omega (20\text{s})=84.0\text{rad}\text{/}\text{s}$;
b. $\tau =3.03\times {10}^3\text{N}·\text{m}$

a. $L=1.131\times {10}^7\text{kg}·{\text{m}}^2\text{/}\text{s}$;
b. $\tau =3.77\times {10}^4\text{N}·\text{m}$

$\omega =28.6\text{rad}\text{/}\text{s}\Rightarrow L=2.6\text{kg}·{\text{m}}^2\text{/}\text{s}$

$L_f=\frac{2}{5}{M}_S{(3.5\text{×}{10}^3\text{km})}^2\frac{2\pi }{T_f}$,
$\begin{array}{}\\ {(7.0\text{×}{10}^5\text{km})}^2\frac{2\pi }{28\text{days}}={(3.5\text{×}{10}^3\text{km})}^2\frac{2\pi }{T_f}\Rightarrow T_f\hfill \\ \\ \\ =28\text{days}\frac{{(3.5\text{×}{10}^3\text{km})}^2}{{(7.0\text{×}{10}^5\text{km})}^2}=7.0\text{×}{10}^{-4}\text{day}=60.5\text{s}\end{array}$

$f_{\text{f}}=2.1\text{rev}\text{/}\text{s}\Rightarrow f_0=0.5\text{rev}\text{/}\text{s}$

$r_{P}m{v}_P=r_{A}m{v}_A\Rightarrow v_P=10.2\text{km}\text{/}\text{s}$

a. $I_{\text{disk}}=5.0\times {10}^{\mathrm{-4}}\text{kg}·{\text{m}}^2$,
$I_{\text{bug}}=2.0\text{×}{10}^{\mathrm{-4}}\text{kg}·{\text{m}}^2$,
$\begin{array}{}\\ (I_{\text{disk}}+I_{\text{bug}}){\omega }_1=I_{\text{disk}}{\omega }_2,\hfill \\ \\ \\ {\omega }_2=14.0\text{rad}\text{/}\text{s}\hfill \end{array}$
b. $\text{Δ}K=0.014\text{J}$;
c. ${\omega }_3=10.0\text{rad}\text{/}\text{s}$ back to the original value;
d. $\frac{1}{2}(I_{\text{disk}}+I_{\text{bug}}){\omega }_3^2=0.035\text{J}$ back to the original value;
e. work of the bug crawling on the disk

$L_{\text{i}}=400.0\text{kg}·{\text{m}}^2\text{/}\text{s}$,
$L_{\text{f}}=500.0\text{kg}·{\text{m}}^2\omega$,
$\omega =0.80\text{rad}\text{/}\text{s}$

$I_0=340.48\text{kg}·{\text{m}}^2$,
$I_{\text{f}}=268.8\text{kg}·{\text{m}}^2$,
${\omega }_{\text{f}}=25.33\text{rpm}$

a. $L=280\text{kg}·{\text{m}}^2\text{/}\text{s}$,
$I_{\text{f}}=89.6\text{kg}·{\text{m}}^2$,
${\omega }_{\text{f}}=3.125\text{rad}\text{/}\text{s}$; b. $K_{\text{i}}=437.5\text{J}$,
$K_{\text{f}}=437.5\text{J}$

Moment of inertia in the record spin: $I_0=0.5\text{kg}·{\text{m}}^2$,
$I_{\text{f}}=1.1\text{kg}·{\text{m}}^2$,
${\omega }_{\text{f}}=\frac{I_0}{I_{\text{f}}}{\omega }_0\Rightarrow f_{\text{f}}=155.5\text{rev}\text{/}\text{min}$

Her spin rate in the air is: $f_{\text{f}}=2.0\text{rev}\text{/}\text{s}$;
She can do four flips in the air.

Moment of inertia with all children aboard:
$I_0=2.4\text{×}{10}^5\text{kg}·{\text{m}}^2$;
$I_{\text{f}}=1.5\text{×}{10}^5\text{kg}·{\text{m}}^2$;
$f_{\text{f}}=0.3\text{rev}\text{/}\text{s}$

$I_0=1.00\text{×}{10}^{10}\text{kg}·{\text{m}}^2$,
$I_{\text{f}}=9.94\text{×}{10}^9\text{kg}·{\text{m}}^2$,
$f_{\text{f}}=3.32\text{rev}\text{/}\text{min}$

$I=2.5\times {10}^{\mathrm{-3}}\text{kg}·{\text{m}}^2$,
${\omega }_P=1.17\text{rad}\text{/}\text{s}$

a. $L_{\text{Earth}}=7.06\text{×}{10}^{33}\text{kg}·{\text{m}}^2\text{/}\text{s}$,
$\text{Δ}L=5.63\text{×}{10}^{33}\text{kg}·{\text{m}}^2\text{/}\text{s}$;
b. $\tau =1.4\text{×}{10}^{22}\text{N}·\text{m}$;
c. The two forces at the equator would have the same magnitude but different directions, one in the north direction and the other in the south direction on the opposite side of Earth. The angle between the forces and the lever arms to the center of Earth is $90\text{°}$, so a given torque would have magnitude $\tau =F{R}_{\text{E}}\text{sin}90\text{°}=F{R}_{\text{E}}$. Both would provide a torque in the same direction:
$\tau =2F{R}_{\text{E}}\Rightarrow F=1.1\text{×}{10}^{15}\text{N}$

$a_{\text{CM}}=-\frac{3}{10}g$,
$v^2=v_0^2+2a_{\text{CM}}x\Rightarrow v^2=(7.0\text{m}\text{/}{\text{s})}^2-2\left(\frac{3}{10}g\right)x,$ $v^2=0\Rightarrow x=8.34\text{m;}$
b. $t=\frac{v-v_0}{a_{\text{CM}}},$ $v=v_0+a_{\text{CM}}t\Rightarrow t=2.38\text{s}$;
The hollow sphere has a larger moment of inertia, and therefore is harder to bring to a rest than the marble, or solid sphere. The distance travelled is larger and the time elapsed is longer.

a. $W=\mathrm{-500.0}\text{J}$;
b. $K+U_{\text{grav}}=\text{constant}$,
$500\text{J}+0=0+(6.0\text{kg})(9.8\text{m}\text{/}{\text{s}}^2)h$,
$h=8.5\text{m,}d=17.0\text{m}$;
The moment of inertia is less for the hollow sphere, therefore less work is required to stop it. Likewise it rolls up the incline a shorter distance than the hoop.

a. $\tau =34.0\text{N}·\text{m}$;
b. $l=m{r}^2\omega \Rightarrow \omega =3.6\text{rad}\text{/}\text{s}$

a. $d_{\text{M}}=3.85\text{×}{10}^8\text{m}$ average distance to the Moon; orbital period $27.32\text{d}=2.36\text{×}{10}^6\text{s}$; speed of the Moon $\frac{2\pi 3.85\text{×}{10}^8\text{m}}{2.36\text{×}{10}^6\text{s}}=1.0\text{×}{10}^3\text{m}\text{/}\text{s}$; mass of the Moon $7.35\text{×}{10}^{22}\text{kg}$,
$L=2.90\text{×}{10}^{34}\text{kg}{\text{m}}^2\text{/}\text{s}$;
b. radius of the Moon $1.74\text{×}{10}^6\text{m}$; the orbital period is the same as (a): $\omega =2.66\text{×}{10}^{\mathrm{-6}}\text{rad}\text{/}\text{s}$,
$L=2.37\text{×}{10}^{29}\text{kg}·{\text{m}}^2\text{/}\text{s}$;
The orbital angular momentum is $1.22\times {10}^5$ times larger than the rotational angular momentum for the Moon.

$I=0.135\text{kg}·{\text{m}}^2$,
$\alpha =4.19\text{rad}\text{/}{\text{s}}^2,$ $\omega ={\omega }_0+\alpha t$,
$\omega (5\text{s})=21.0\text{rad}\text{/}\text{s},$ $L=2.84\text{kg}·{\text{m}}^2\text{/}\text{s}$,
$\omega (10\text{s})=41.9\text{rad}\text{/}\text{s},$ $L=5.66\text{kg}·\text{m}\text{/}{\text{s}}^2$

In the conservation of angular momentum equation, the rotation rate appears on both sides so we keep the (rev/min) notation as the angular velocity can be multiplied by a constant to get (rev/min):
$\begin{array}{ccc}\hfill L_{\text{i}}& =\hfill & \mathrm{-0.04}\text{kg}·{\text{m}}^2(300.0\text{rev}\text{/}\text{min),}\hfill \\ \hfill L_{\text{f}}& =\hfill & 0.08\text{kg}·{\text{m}}^2{f}_{\text{f}}\Rightarrow f_{\text{f}}=\mathrm{-150.0}\text{rev}\text{/}\text{min clockwise}\hfill \end{array}$

$I_0{\omega }_0=I_{\text{f}}{\omega }_{\text{f}}$,
$I_0=6120.0\text{kg}·{\text{m}}^2$,
$I_{\text{f}}=1180.0\text{kg}·{\text{m}}^2$,
${\omega }_{\text{f}}=31.1\text{rev}\text{/}\text{min}$

$L_{\text{i}}=1.00\text{×}{10}^7\text{kg}·{\text{m}}^2\text{/}\text{s}$,
$I_{\text{f}}=2.025\text{×}{10}^5\text{kg}·{\text{m}}^2$,
${\omega }_{\text{f}}=7.86\text{rev}\text{/}\text{s}$

Assume the roll accelerates forward with respect to the ground with an acceleration $a^{\prime }$. Then it accelerates backwards relative to the truck with an acceleration $(a-a^{\prime })$.

Also, $R\alpha =a-a^{\prime }I=\frac{1}{2}m{R}^2{\displaystyle \sum F_x=f_s=m{a}^{\prime }}$,
${\displaystyle \sum \tau }=f_{s}R=I\alpha =I\frac{a-a^{\prime }}{R}{f}_s=\frac{I}{R^2}(a-a^{\prime })=\frac{1}{2}m(a-a^{\prime })$,
Solving for $a^{\prime }$: $f_s=\frac{1}{2}m(a-a^{\prime })$; $a^{\prime }=\frac{a}{3}$,
$x-x_0=v_{0}t+\frac{1}{2}a{t}^2;d=\frac{1}{3}a{t}^2;t=\sqrt{\frac{3d}{a}};$
therefore, $s=1.5d$

a. The tension in the string provides the centripetal force such that $T\text{sin}\theta =m{r}_{\perp }{\omega }^2$. The component of the tension that is vertical opposes the gravitational force such that $T\text{cos}\theta =mg$. This gives $T=5.7\text{N}$. We solve for $r_{\perp }=0.16\text{m}$. This gives the length of the string as $r=0.32\text{m}$.
At $\omega =10.0\text{rad}\text{/}\text{s}$, there is a new angle, tension, and perpendicular radius to the rod. Dividing the two equations involving the tension to eliminate it, we have $\frac{\text{sin}\theta }{\text{cos}\theta }=\frac{(0.32\text{m}\text{sin}\theta ){\omega }^2}{g}\Rightarrow \frac{1}{\text{cos}\theta }=\frac{0.32\text{m}{\omega }^2}{g}$;
$\text{cos}\theta =0.31\Rightarrow \theta =72.2\text{°}$; b. $l_{\text{initial}}=0.08\text{kg}·{\text{m}}^2\text{/}\text{s}$,
$l_{\text{final}}=0.46\text{kg}·{\text{m}}^2\text{/}\text{s}$; c. No, the cosine of the angle is inversely proportional to the square of the angular velocity, therefore in order for $\theta \to 90\text{°},\omega \to \infty$. The rod would have to spin infinitely fast.