Respuesta :
With constant angular acceleration [tex]\alpha[/tex], the disk achieves an angular velocity [tex]\omega[/tex] at time [tex]t[/tex] according to
[tex]\omega=\alpha t[/tex]
and angular displacement [tex]\theta[/tex] according to
[tex]\theta=\dfrac12\alpha t^2[/tex]
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
[tex]21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}[/tex]
b. Under constant acceleration, the average angular velocity is equivalent to
[tex]\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2[/tex]
where [tex]\omega_f[/tex] and [tex]\omega_i[/tex] are the final and initial angular velocities, respectively. Then
[tex]\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}[/tex]
c. After 1.00 s, the disk has instantaneous angular velocity
[tex]\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}[/tex]
d. During the next 1.00 s, the disk will start moving with the angular velocity [tex]\omega_0[/tex] equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle [tex]\theta[/tex] according to
[tex]\theta=\omega_0t+\dfrac12\alpha t^2[/tex]
which would be equal to
[tex]\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}[/tex]