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$$
\begin{array}{l}{\text { A woman with mass } 50 \mathrm{kg} \text { is standing on the rim of a }} \\ {\text { large disk that is rotating at } 0.50 \text { rev/s about an axis through its }} \\ {\text { center. The disk has mass } 110 \mathrm{kg} \text { and radius } 4.0 \mathrm{m} \text { . Calculate the }} \\ {\text { magnitude of the total angular momentum of the woman-disk }} \\ {\text { system. (Assume that you can treat the woman as a point.) }}\end{array}
$$

$$
\omega=0.5 \mathrm{rev} / \mathrm{s}=3.14  \text { rad/s }
$$

$$
I=I_{d i s k}+I_{\text {woman }}
$$

$$
I_{disk} \rightarrow \frac{1}{2} M_{disk}R^2
$$

$$
I_\text { woman } \rightarrow M_{woman} R^2
$$

$$
L=I \omega
$$

$$
I=R^{2}\left(\frac{1}{2} M_{disk}+M_{woman}\right)=(4)^{2}\left(\frac{1}{2}(110+50)\right)=1680 \mathrm{kg} \cdot \mathrm{m}^{2}
$$

$$
L=(1680)(3.14)=5.28 * 10^{3} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}
$$

$$
\begin{array}{l}{\text { Find the magnitude of the angular momentum of the }} \\ {\text { second hand on a clock about an axis through the center of the }} \\ {\text { clock face. The clock hand has a length of } 15.0 \mathrm{cm} \text { and a mass of }} \\ {6.00 \text { g. Take the second hand to be a slender rod rotating with con- }} \\ {\text { stant angular velocity about one end. }}\end{array}
$$

$$
L=I \omega
$$

$$
{\omega=1 \text { rev/min }(2 \pi \mathrm{rad} / \mathrm{1rev})(1min / 60 \mathrm{s})=0.1047 \mathrm{rad/s}}
$$

$$
I=\frac{1}{3} M L^{2}=\frac{1}{3}\left(6 * 10^{-3}\right)(0.15)^{2}=4.5 * 10^{-5} kg  \cdot m^{2}
$$

$$
\longrightarrow L=I w
$$

$$
=4.5 * 10^{-5} * 0.1047=4.71*10^{-6} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}
$$

$$
\vec{L} \ \mathrm{c.w}
$$

$$
\begin{array}{l}{\text { A hollow, thin-walled sphere of mass } 12.0 \mathrm{kg} \text { and }} \\ {\text { diameter } 48.0 \mathrm{cm} \text { is rotating about an axle through its conter. The }} \\ {\text { angle (in radians) through which it turns as a function of time (in }} \\ {\text { seconds) is given by } \theta(t)=A t^{2}+B t^{4}, \text { where } A \text { has numerical }} \\ {\text { value } 1.50 \text { and } B \text { has numerical value } 1.10 . \text { (a) What are the units }} \\ {\text { of the constants } A \text { and } B ?(\text { b) At the time } 3.00 \text { s, find (i) the angular }} \\ {\text { momentum of the sphere and (ii) the net torque on the sphere. }}\end{array}
$$

$$
I=\frac{2}{3} M R^{2}
$$

$$
R=0.24 \mathrm{m}
$$

$$
\theta(t)=1.5 t^{2}+1.1 t^{4} \rightarrow A,B \quad radians
$$

$$
\omega_{z}=\frac{d \theta}{dt}=2 At+4 B t^{3}
$$

$$
\omega_{z}=2(1.5)(3)+4(1.1)(3)^{3}=128 \mathrm{rad} / \mathrm{s}
$$

$$
L_{z} =\left(\frac{2}{3} M R^{2}\right) \omega_{z}=\frac{2}{3}(12)(0.24)^{2}(128)=59 \quad kg \cdot m^{2} / s
$$

$$
\tau_{z}=\frac{d L_ z}{d t}=I \frac{d \omega_ z}{d t}=I\left(2 A+12 B t^{2}\right)
$$

$$
\tau_ z=\frac{2}{3}(12)(0.24)^2(2(1.5)+12(1.1)(3)^2)=56.1 N.m
$$

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