Respuesta :
Answer:
(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Explanation:
Given that,
Initial spinning = 50.0 rad/s
Time = 20.0
Distance = 2.5 m
Mass of pole = 4 kg
Angle = 60°
We need to calculate the angular acceleration
Using formula of angular velocity
[tex]\omega=-\alpha t[/tex]
[tex]\alpha=-\dfrac{\omega}{t}[/tex]
[tex]\alpha=-\dfrac{50.0}{20.0}[/tex]
[tex]\alpha=-2.5\ rad/s^2[/tex]
The angular acceleration is -2.5 rad/s²
We need to calculate the number of revolution
Using angular equation of motion
[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t[/tex]
Put the value into the formula
[tex]\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2[/tex]
[tex]\theta=500\ rad[/tex]
The number of revolution is 500 rad.
(II). We need to calculate the torque
Using formula of torque
[tex]\vec{\tau}=\vec{r}\times\vec{f}[/tex]
[tex]\tau=r\times f\sin\theta[/tex]
Put the value into the formula
[tex]\tau=2.5\times4\times 9.8\sin60[/tex]
[tex]\tau=84.87\ N-m[/tex]
Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
