Answer:
0.321659377 m/s²
1.383138458 m/s
0.321659377 m/s²
0.62667 m/s²
0.7044 m/s² and 27.17°
Explanation:
d = Diameter of rim = 13 in = [tex]13\times 0.0254=0.3302\ m[/tex]
r = Radius = [tex]\frac{d}{2}=\frac{0.3302}{2}=0.1651\ m[/tex]
[tex]\omega_f[/tex] = Final angular velocity = [tex]80\times\frac{2\pi}{60}=8.37758\ rad/s[/tex]
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\alpha[/tex] = Angular acceleration
t = Time taken = 4.3 s
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2[/tex]
Tangential acceleration is given by
[tex]a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2[/tex]
The tangential acceleration of the bug is 0.321659377 m/s²
Tangential velocity is given by
[tex]v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s[/tex]
The tangential velocity of the bug is 1.383138458 m/s
The tangential acceleration is constant which is 0.321659377 m/s²
Centripetal acceleration is given by
[tex]a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2[/tex]
The centripetal acceleration of the bug is 0.62667 m/s²
The resultant of the acceleration gives us total acceleration
[tex]a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2[/tex]
Direction is given by
[tex]\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}[/tex]
The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°
