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$$
\begin{array}{l}{\text { (a) What angle in radians is subtended by an arc } 1.50 \mathrm{m} \text { long }} \\ {\text { on the circumference of a circle of radius } 2.50 \mathrm{m} ? \text { What is this }} \\ {\text { angle in degrees? (b) An arc } 14.0 \mathrm{cm} \text { long on the circumference of }} \\ {\text { a circle subtends an angle of } 128^{\circ} . \text { What is the radius of the circle? }} \\ {\text { (c) The angle between two radii of a circle with radius } 1.50 \mathrm{m} \text { is }} \\ {0.700 \text { rad. What length of arc is intercepted on the circumference }} \\ {\text { of the circle by the two radii? }}\end{array}
$$

$$
S=r \theta
$$

$$
\pi \ rad=180^{\circ} \longrightarrow \theta=\frac{S}{r}
$$

(a) $$
\theta=\frac{1.5}{2.5}=0.6 \mathrm{rad}=34.4^{\circ}
$$

(b) $$
r=\frac{s}{\theta}=\frac{14}{128^{\circ}\left(\frac{\pi \ rad}{180^{\circ}}\right)}=6.27 \mathrm{cm}
$$

(c) $$
S=r \theta=1.5 * 0.7=1.05 \mathrm{m}
$$

$$
\begin{array}{l}{\text { A fan blade rotates with angular velocity given by }} \\ {\omega_{z}(t)=\gamma-\beta t^{2}, \text { where } \gamma=5.00 \mathrm{rad} / \mathrm{s} \text { and } \beta=0.800 \mathrm{rad} / \mathrm{s}^{3} .} \\ {\text { (a) Calculate the angular acceleration as a function of time. }} \\ {\text { (b) Calculate the instantaneous angular acceleration } \alpha_{z} \text { at } t=3.00 \mathrm{s}} \\ {\text { and the average angular acceleration } \alpha_{3 v-2} \text { for the time interval }} \\ {t=0 \text { to } t=3.00 \mathrm{s} \text { . How do these two quantities compare? If they }} \\ {\text { are different, why are they different? }}\end{array}
$$

$$
\alpha_{z}=\frac{d w_{z}}{d t} \quad, \quad \alpha_{a v-z}=\frac{\Delta w_{z}}{\Delta t}
$$

(a) $$
\alpha_{z}(t)=\frac{d w_{z}}{d t}=-2 \beta t=\left(-1.6 \mathrm{rad} / \mathrm{s}^{3}\right) t
$$

(b) $$
\alpha_{z}(3)=(-1.6)(3)=-4.8 \mathrm{rad} / \mathrm{s}^{2}
$$

$$
\alpha_{a v-z}=\frac{\Delta w_{z}(0 \rightarrow 3)}{\Delta t}=\frac{w_{z}(3)-w_{z} (0)}{3-0}
$$

$$
=\frac{-2 .2-5}{3}=-2.4 \mathrm{rad} / \mathrm{s}^{2}
$$

$$
\begin{array}{l}{\text { At } t=0 \text { the current to a de electric motor is reversed, }} \\ {\text { resulting in an angular displacement of the motor shaft given by }} \\ {\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} . \text { (a) At }} \\ {\text { what time is the angular velocity of the motor shaft zero? }} \\ {\text { (b) Calculate the angular acceleration at the instant that the motor }} \\ {\text { shaft has zero angular velocity. (c) How many revolutions does the }}\end{array}
$$

$$
\begin{array}{l}{\text { motor shaft turn through between the time when the current is }} \\ {\text { reversed and the instant when the angular velocity is zero? (d) How }} \\ {\text { fast was the motor shaft rotating at } t=0, \text { when the current was }} \\ {\text { reversed? (e) Calculate the average angular velocity for the time }} \\ {\text { period from } t=0 \text { to the time calculated in part (a). }}\end{array}
$$

$$
w_z (t)=\frac{d \theta}{d t} \quad, \quad \alpha_{z}(t)=\frac{d w_ z}{d t} \quad, \quad w_{a v-z}=\frac{\Delta \theta}{\Delta t}
$$

(a) $$
w_{z}=0 \quad \longrightarrow \quad 250-40 t-4.5 t^{2}=0
$$

\(∴ t=4.23 \mathrm{s} \)

(b) $$
@ \ t=4.23 \mathrm{s} \rightarrow \alpha_{z}=-78.1 \mathrm{rad} / \mathrm{s}^{2}
$$

(c) $$
@ \ t=4.23s \rightarrow \theta=586 \mathrm{rad}=93.3 \mathrm{rev.}
$$

(d) $$
@ \ t=0 \rightarrow w_{z}=250-0-0=250 \mathrm{rad} / \mathrm{s}
$$

(e) $$
w_{a v-z}=\frac{\Delta \theta}{\Delta t}=\frac{586}{4.23}=138 \mathrm{rad/s}
$$

 

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