Givens.
• R = 30 cm.
,• t = 4 seconds.
,• a_n = 2.7 m/s^2.
,• n = 3 revolutions.
First, we need to find the angular acceleration.
[tex]\theta=\omega_ot+\frac{1}{2}\varepsilon t^2[/tex]But, we need to find the angular displacement and the angular speed.
[tex]\begin{gathered} \theta=3rev\cdot\frac{2\pi}{1rev}\cdot=6\pi \\ a_n=\frac{v^2}{R}\to v^2=a_n\cdot R=v=\sqrt[]{2.7\cdot\frac{m}{s^2}\cdot0.30m} \\ v=0.9\cdot\frac{m}{s} \end{gathered}[/tex]The displacement of 3 revolutions is 6pi radians and the tangential (linear) speed of 0.9 m/s. Use the radius to find the angular speed.
[tex]\begin{gathered} \omega=\frac{v}{R}=\frac{0.9\cdot\frac{m}{s}}{0.30m} \\ \omega=3\cdot\frac{\text{rad}}{\sec } \end{gathered}[/tex]Once we have the angular speed and the angular displacement, we are able to find the angular acceleration.
[tex]\begin{gathered} \theta=\omega_ot+\frac{1}{2}\varepsilon t^2 \\ 6\pi=3\cdot4+\frac{1}{2}\cdot\varepsilon(4)^2 \\ 6\pi=12+6\varepsilon \\ \varepsilon=\frac{6\pi-12}{6} \\ \varepsilon\approx1.14\cdot\frac{rad}{\sec ^2} \end{gathered}[/tex]Once we have the angular acceleration, we can find the tangential acceleration.
[tex]a_t=\varepsilon R=1.14\cdot0.30\approx0.34\cdot\frac{m}{s^2}[/tex]The tangential acceleration is 0.34 m/s^2.
The total acceleration would be
[tex]\begin{gathered} a_{\text{total}}=\sqrt[]{a^2_t+a^2_n}=\sqrt[]{(0.34)^2+(2.7)^2} \\ a_{\text{total}}=\sqrt[]{0.1156+7.29}\approx2.72\cdot\frac{m}{s^2} \end{gathered}[/tex]The total acceleration is 2.72 m/s^2.
The linear velocity is 0.9 m/s.
The angular velocity is 3 rad/sec.
