Answer:
Explanation:
Given that:
the initial angular velocity [tex]\omega_o = 0[/tex]
angular acceleration [tex]\alpha[/tex] = 4.44 rad/s²
Using the formula:
[tex]\omega = \omega_o+ \alpha t[/tex]
Making t the subject of the formula:
[tex]t= \dfrac{\omega- \omega_o}{ \alpha }[/tex]
where;
[tex]\omega = 1.53 \ rad/s^2[/tex]
∴
[tex]t= \dfrac{1.53-0}{4.44 }[/tex]
t = 0.345 s
b)
Using the formula:
[tex]\omega ^2 = \omega _o^2 + 2 \alpha \theta[/tex]
here;
[tex]\theta[/tex] = angular displacement
∴
[tex]\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }[/tex]
[tex]\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }[/tex]
[tex]\theta =0.264 \ rad[/tex]
Recall that:
2π rad = 1 revolution
Then;
0.264 rad = (x) revolution
[tex]x = \dfrac{0.264 \times 1}{2 \pi}[/tex]
x = 0.042 revolutions
c)
Here; force = 270 N
radius = 1.20 m
The torque = F * r
[tex]\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm[/tex]
However;
From the moment of inertia;
[tex]Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}[/tex]
given that;
I = 84.4 kg.m²
[tex]\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2[/tex]
For re-tardation; [tex]\alpha=-3.84 \ rad/s^2[/tex]
Using the equation
[tex]t= \dfrac{\omega- \omega_o}{ \alpha }[/tex]
[tex]t= \dfrac{0-1.53}{ -3.84 }[/tex]
[tex]t= \dfrac{1.53}{ 3.84 }[/tex]
t = 0.398s
The required time it takes= 0.398s
