Complete Question
The complete is shown on the first uploaded image
Answer:
[tex]\alpha = 5.89 rad/s^2[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
[tex]v= 0.212m/s[/tex]
[tex]a_t= 0.5301 m/s[/tex]
[tex]a_r = 0.499 m/s[/tex]
[tex]a = 0.7279 m/s[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
Explanation:
From the question we are told that
mass of the ant is [tex]m_a = 0.8g = \frac{0.8}{1000} = 0.00018kg[/tex]
The distance from the center is [tex]d = 9cm = \frac{9}{100} = 0.09m[/tex]
The angular speed is [tex]w = 45rpm = 45 * \frac{2 \pi }{60} = 1.5 \pi[/tex]
The time taken to attain angular acceleration of 45rpm [tex]t_1 = 0.8s[/tex]
The time taken is [tex]t_2 = 0.4 s[/tex]
The angular acceleration is mathematically represented as
[tex]\alpha = \frac{w}{t}[/tex]
[tex]= \frac{1.5}{0.8}[/tex]
[tex]\alpha = 5.89 rad/s^2[/tex]
The angular velocity at time t= 0.4s is mathematically represented as
[tex]w__{0.4s}} = \alpha * t_2[/tex] Recall angular acceleration is constant
[tex]= 5.89 * 0.4[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
The linear velocity is mathematically represented as
[tex]v = w__{t_2}} * r[/tex]
[tex]= 2.36 * 0.09[/tex]
[tex]v= 0.212m/s[/tex]
The tangential acceleration is mathematically represented as
[tex]a_{t} = \alpha * r[/tex]
[tex]= 5.89 * 0.09[/tex]
[tex]a_t= 0.5301 m/s[/tex]
The radial acceleration is mathematically represented as
[tex]a_r = \frac{v^2}{r}[/tex]
[tex]= \frac{0.212^2}{0.09}[/tex]
[tex]a_r = 0.499 m/s[/tex]
The resultant velocity is mathematically represented as
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
[tex]= \sqrt{0.53^2 + 0.499^2}[/tex]
[tex]a = 0.7279 m/s[/tex]
The net force is mathematically represented as
[tex]F_{net} = 0.0008 * 0.7279[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
