Answer:
a
The angular acceleration is [tex]\alpha = 0.808 rad/s^2[/tex]
b
The average angular velocity is [tex]w = 2.02 \ rad/s[/tex]
c
The instantaneous angular velocity is [tex]w = 4.04 \ rad/s[/tex]
d
The additional angle (rad) will the disk turn during the next 5.00 s is
[tex]\theta_n = 30.30 \ rad[/tex]
Explanation:
From the question we are told that
The duration of rotation is [tex]t = 5.00s[/tex]
The angular displacement is [tex]\theta= 10.1 rad[/tex]
From newtons Law the angular displacement is mathematically represented as
[tex]\theta = w_o t +\frac{1}{2} \alpha t^2[/tex]
Where [tex]\alpha[/tex] is the angular acceleration
[tex]w_o[/tex] is the initial angular velocity and its value is 0
Making [tex]\alpha[/tex] the subject of formula
[tex]\alpha = \frac{\theta}{0.5 * t^2}[/tex]
Substituting value
[tex]\alpha = \frac{10.1}{0.5 * 5^2 }[/tex]
[tex]\alpha = 0.808 rad/s^2[/tex]
The average angular velocity is mathematically represented as
[tex]w = \frac{\theta }{t}[/tex]
Substituting value
[tex]w = \frac{10.1}{5}[/tex]
[tex]w = 2.02 \ rad/s[/tex]
From the equations of motion the instantaneous angular velocity is mathematically represented as
[tex]w = w_o + \alpha t[/tex]
Substituting value
[tex]w = 0 + 0.808 * 5[/tex]
[tex]w = 4.04 \ rad/s[/tex]
Modifying the equation for angular displacement is mathematically represented as
[tex]\theta_{n} = w_o t + 0.5 \alpha t^2[/tex]
From the question [tex]\alpha[/tex] is not changed
t = 5s
the initial velocity would be the instantaneous velocity
So [tex]w_o = 4.04 \ rad/s[/tex]
Substituting value
[tex]\theta_n = 4.04 * 5 + (0.5 * 0.808 * 5^2)[/tex]
[tex]\theta_n = 30.30 \ rad[/tex]
