Chapter 10

a. $40.0\text{rev}\text{/}\text{s}=2\pi (40.0)\text{rad}\text{/}\text{s}$, $\bar{\alpha }=\frac{\text{Δ}\omega }{\text{Δ}t}=\frac{2\pi (40.0)-0\text{rad/s}}{20.0\text{s}}=2\pi (2.0)=4.0\pi \text{rad/}{\text{s}}^2$; b. Since the angular velocity increases linearly, there has to be a constant acceleration throughout the indicated time. Therefore, the instantaneous angular acceleration at any time is the solution to $4.0\pi \text{rad/}{\text{s}}^2$.

a. Using Equation 10.11, we have $7000\text{rpm}=\frac{7000.0(2\pi \text{rad})}{60.0\text{s}}=733.0\text{rad}\text{/}\text{s},$
$\alpha =\frac{(\omega -{\omega }_0)}{t}=\frac{(0–733.0\text{rad}\text{/}\text{s})}{10.0\text{s}}=\mathrm{–73.3}\text{rad}\text{/}{\text{s}}^2$;
b. Using Equation 10.13, we have
${\omega }^2={\omega }_0^2+2\alpha \text{Δ}\theta \Rightarrow \text{Δ}\theta =\frac{{\omega }^2-{\omega }_0^2}{2\alpha }=\frac{0-{(733.0\text{rad}\text{/}\text{s})}^2}{2(\mathrm{–73.3}\text{rad}\text{/}{\text{s}}^2)}=3665.2\text{rad}$

The angular acceleration is $\alpha =\frac{(5.0-0)\text{rad}\text{/}\text{s}}{20.0\text{s}}=0.25\text{rad}\text{/}{\text{s}}^2$. Therefore, the total angle that the boy passes through is
$\text{Δ}\theta =\frac{{\omega }^2-{\omega }_0^2}{2\alpha }=\frac{{(5.0)}^2-0}{2(0.25)}=50\text{rad}$.
Thus, we calculate
$s=r\theta =5.0\text{m}(50.0\text{rad})=250.0\text{m}$.

The initial rotational kinetic energy of the propeller is
$K_0=\frac{1}{2}I{\omega }^2=\frac{1}{2}(800.0{\text{kg-m}}^2)(4.0\times 2\pi \text{rad}\text{/}{\text{s})}^2=2.53\times {10}^5\text{J}$.
At 5.0 s the new rotational kinetic energy of the propeller is
$K_{\text{f}}=2.03\times {10}^5\text{J}$.
and the new angular velocity is
$\omega =\sqrt{\frac{2(2.03\times {10}^5\text{J})}{800.0{\text{kg-m}}^2}}=22.53\text{rad}\text{/}\text{s}$
which is 3.58 rev/s.

$I_{\text{parallel-axis}}=I_{\text{center of mass}}+m{d}^2=m{R}^2+m{R}^2=\left(\frac{1}{3}\right)m{R}^2$

The angle between the lever arm and the force vector is $80\text{°};$ therefore, $r_{\perp }=100\text{m(sin80}\text{°})=98.5\text{m}$.

The cross product $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$ gives a negative or clockwise torque.

The torque is then $\tau =\text{−}{r}_{\perp }F=\mathrm{-98.5}\text{m}(5.0\times {10}^5\text{N})=\mathrm{-4.9}\times {10}^7\text{N}·\text{m}$.

a. The angular acceleration is $\alpha =\frac{20.0(2\pi )\text{rad}\text{/}\text{s}-0}{10.0\text{s}}=12.56\text{rad}\text{/}{\text{s}}^2$. Solving for the torque, we have ${\displaystyle \sum _i{\tau }_i}=I\alpha =(30.0\text{kg}·{\text{m}}^2)(12.56\text{rad}\text{/}{\text{s}}^2)=376.80\text{N}·\text{m}$; b. The angular acceleration is $\alpha =\frac{0-20.0(2\pi )\text{rad}\text{/}\text{s}}{20.0\text{s}}=\mathrm{-6.28}\text{rad}\text{/}{\text{s}}^2$. Solving for the torque, we have ${\displaystyle \sum _i{\tau }_i}=I\alpha =(30.0{\text{kg-m}}^2)(\mathrm{-6.28}\text{rad}\text{/}{\text{s}}^2)=\mathrm{-188.50}\text{N}·\text{m}$

3 MW

The second hand rotates clockwise, so by the right-hand rule, the angular velocity vector is into the wall.

They have the same angular velocity. Points further out on the bat have greater tangential speeds.

straight line, linear in time variable

constant

The centripetal acceleration vector is perpendicular to the velocity vector.

a. both; b. nonzero centripetal acceleration; c. both

The hollow sphere, since the mass is distributed further away from the rotation axis.

a. It decreases. b. The arms could be approximated with rods and the discus with a disk. The torso is near the axis of rotation so it doesn’t contribute much to the moment of inertia.

Because the moment of inertia varies as the square of the distance to the axis of rotation. The mass of the rod located at distances greater than L/2 would provide the larger contribution to make its moment of inertia greater than the point mass at L/2.

magnitude of the force, length of the lever arm, and angle of the lever arm and force vector

The moment of inertia of the wheels is reduced, so a smaller torque is needed to accelerate them.

yes

$\left|\overrightarrow{r}\right|$ can be equal to the lever arm but never less than the lever arm

If the forces are along the axis of rotation, or if they have the same lever arm and are applied at a point on the rod.

$\omega =\frac{2\pi \text{rad}}{45.0\text{s}}=0.14\text{rad/s}$

a. $\theta =\frac{s}{r}=\frac{3.0\text{m}}{1.5\text{m}}=2.0\text{rad}$; b. $\omega =\frac{2.0\text{rad}}{1.0\text{s}}=2.0\text{rad/s}$; c. $\frac{v^2}{r}=\frac{{(3.0\text{m/s})}^2}{1.5\text{m}}=6.0\text{m}\text{/}{\text{s}}^2.$

The propeller takes only $\text{Δ}t=\frac{\text{Δ}\omega }{\alpha }=\frac{0\text{rad/s}-10.0(2\pi )\text{rad/s}}{\mathrm{-2.0}\text{rad}\text{/}{\text{s}}^2}=31.4\text{s}$ to come to rest, when the propeller is at 0 rad/s, it would start rotating in the opposite direction. This would be impossible due to the magnitude of forces involved in getting the propeller to stop and start rotating in the opposite direction.

a. $\omega =25.0(2.0\text{s})=50.0\text{rad/s}$; b. $\alpha =\frac{d\omega }{dt}=25.0{\text{rad/s}}^2$

a. $\omega =55\text{rad}\text{/}\text{s}$;
b. $t=11\text{s}$

a. $0.87\text{rad}\text{/}{\text{s}}^2$;
b. $\theta =\mathrm{12,600}\text{rad}$

a. $\omega =42.0\text{rad}\text{/}\text{s}$;
b. $\theta =220\text{rad}$; c. $\begin{array}{}\\ \\ v_t=42\text{m}\text{/}\text{s}\hfill \\ a_t=4.0\text{m}\text{/}{\text{s}}^2\hfill \end{array}$

a. $\omega =7.0\text{rad}\text{/}\text{s}$;
b. $\theta =22.5\text{rad}$; c. $a_{\text{t}}=0.10\text{m}\text{/}{\text{s}}^2$

$\alpha =28.6\text{rad}\text{/}{\text{s}}^2$.

$r=0.78\text{m}$

a. $\alpha =\mathrm{-0.314}\text{rad}\text{/}{\text{s}}^2$,
b. $a_{\text{c}}=197.4\text{m}\text{/}{\text{s}}^2$; c. $a=\sqrt{a_{\text{c}}^2+a_{\text{t}}^2}=\sqrt{{197.4}^2+{(\mathrm{-6.28})}^2}=197.5\text{m}\text{/}{\text{s}}^2$
$\theta ={\text{tan}}^{\mathrm{-1}}\frac{\mathrm{-6.28}}{197.4}=\mathrm{-1.8}\text{°}$ in the counterclockwise direction from the centripetal acceleration vector

$ma=40.0\text{kg}(4.75\text{m}\text{/}{\text{s}}^2)=190\text{N}$
The maximum friction force is ${\mu }_{\text{S}}N=0.6(40.0\text{kg})(9.8\text{m}\text{/}{\text{s}}^2)=235.2\text{N}$ so the child does not fall off yet.

$\begin{array}{ccc}\hfill v_{\text{t}}& =\hfill & r\omega =1.0(2.0t)\text{m/s}\hfill \\ \hfill a_{\text{c}}& =\hfill & \frac{v_t^2}{r}=\frac{{(2.0t)}^2}{1.0\text{m}}=4.0t^2\text{m}\text{/}{\text{s}}^2\hfill \\ \hfill a_{\text{t}}(t)& =\hfill & r\alpha (t)=r\frac{d\omega }{dt}=1.0\text{m}(2.0)=2.0\text{m}\text{/}{\text{s}}^2.\hfill \end{array}$
Plotting both accelerations gives

The tangential acceleration is constant, while the centripetal acceleration is time dependent, and increases with time to values much greater than the tangential acceleration after t = 1s. For times less than 0.7 s and approaching zero the centripetal acceleration is much less than the tangential acceleration.

a. $K=2.56\times {10}^{29}\text{J;}$
b. $K=2.68\times {10}^{33}\text{J}$

$K=434.0\text{J}$

a. $v_{\text{f}}=86.5\text{m}\text{/}\text{s}$;
b. The rotational rate of the propeller stays the same at 20 rev/s.

$K=3.95\times {10}^{42}\text{J}$

a. $I=0.315\text{kg}·{\text{m}}^2$;
b. $K=621.8\text{J}$

$I=\frac{7}{36}m{L}^2$

$v=7.14\text{m}\text{/}\text{s}.$

$\theta =10.2\text{°}$

$F=30\text{N}$

a. $0.85\text{m}\left(55.0\text{N}\right)=46.75\text{N}·\text{m}$; b. It does not matter at what height you push.

$m_2=\frac{4.9\text{N}·\text{m}}{9.8(0.3\text{m})}=1.67\text{kg}$

${\tau }_{net}=\mathrm{-9.0}\text{N}·\text{m}+3.46\text{N}·\text{m}+0-3.38\text{N}·\text{m}=\mathrm{-8.92}\text{N}·\text{m}$

$\tau =5.66\text{N}·\text{m}$

${\displaystyle \sum \tau =}57.82\text{N}·\text{m}$

$\overrightarrow{r}\times \overrightarrow{F}=4.0\widehat{i}+2.0\widehat{j}-16.0\widehat{k}\text{N}·\text{m}$

a. $\tau =(0.280\text{m})(180.0\text{N})=50.4\text{N}·\text{m}$; b. $\alpha =17.14\text{rad}\text{/}{\text{s}}^2$;
c. $\alpha =17.04\text{rad}\text{/}{\text{s}}^2$

$\tau =8.0\text{N}·\text{m}$

$\tau =\mathrm{-43.6}\text{N}·\text{m}$

a. $\alpha =1.4\times {10}^{\mathrm{-10}}\text{rad}\text{/}{\text{s}}^2$;
b. $\tau =1.36\times {10}^{28}\text{N-m}$; c. $F=2.1\times {10}^{21}\text{N}$

$a=3.6\text{m}\text{/}{\text{s}}^2$

a. $a=r\alpha =14.7\text{m}\text{/}{\text{s}}^2$; b. $a=\frac{L}{2}\alpha =\frac{3}{4}g$

$\tau =\frac{P}{\omega }=\frac{2.0\times {10}^6\text{W}}{2.1\text{rad}\text{/}\text{s}}=9.5\times {10}^5\text{N}·\text{m}$

a. $K=888.50\text{J}$;
b. $\text{Δ}\theta =294.6\text{rev}$

a. $I=114.6\text{kg}·{\text{m}}^2$;
b. $P=\mathrm{104,700}\text{W}$

$v=L\omega =\sqrt{3Lg}$

a. $a=5.0\text{m}\text{/}{\text{s}}^2$; b. $W=1.25\text{N}·\text{m}$

$\text{Δ}t=10.0\text{s}$

a. $0.06\text{rad}\text{/}{\text{s}}^2$; b. $\theta =16.7\text{revolutions}$

$s=405.26\text{m}$

a. $I=0.363\text{kg}·{\text{m}}^2$;
b. $I=2.34\text{kg}·{\text{m}}^2$

$\omega =\sqrt{\frac{6.68\text{J}}{4.4{\text{kgm}}^2}}=1.23\text{rad}\text{/}\text{s}$

$F=23.3\text{N}$

$\alpha =\frac{190.0\text{N-m}}{2.94{\text{kg-m}}^2}=64.4{\text{rad}\text{/}\text{s}}^2$

a. $\omega =2.0t-1.5t^2$; b. $\theta =t^2-0.5t^3$; c. $\theta =\mathrm{-400.0}\text{rad}$; d. the vector is at $\mathrm{-0.66}(360\text{°})=\mathrm{-237.6}\text{°}$

$I=\frac{2}{5}m{R}^2$

a. $\omega =8.2\text{rad}\text{/}\text{s}$; b. $\omega =8.0\text{rad}\text{/}\text{s}$