Answer : The angular acceleration is
![\alpha=-\beta t=-0.810\text{ t}](https://img.qammunity.org/2019/formulas/chemistry/high-school/pfir0shafutdn8ftbrhf19zflehgom469w.png)
Solution : Given,
![\gamma=5.20rad/s](https://img.qammunity.org/2019/formulas/chemistry/high-school/4kh2w51ftmxcdmf71g6px8482o7j5diy9m.png)
![\beta=0.810rad/s^3](https://img.qammunity.org/2019/formulas/chemistry/high-school/xeiinrtlrbfz1y8kgdwp91441rm25brgnw.png)
The given angular velocity equation is,
![\omega_z(t)=\gamma-\beta t^2](https://img.qammunity.org/2019/formulas/chemistry/high-school/yulw9dq87ztuwq2oc46lqvneqpj6wxbu3j.png)
At t = 0,
![\omega_z(0)=\gamma](https://img.qammunity.org/2019/formulas/chemistry/high-school/fratoruagx69n9rmgnq3nd5l42590ng58b.png)
At t = t,
![\omega_z(t)=\gamma-\beta t^2](https://img.qammunity.org/2019/formulas/chemistry/high-school/yulw9dq87ztuwq2oc46lqvneqpj6wxbu3j.png)
Formula used for angular acceleration :
![\alpha=(\omega_z(t)-\omega_z(0))/(t)](https://img.qammunity.org/2019/formulas/chemistry/high-school/8d10fk1w6needo53x6zoit1qdv9iitqf3g.png)
where,
= angular acceleration
= angular velocity at time 't'
= angular velocity at time '0'
t = time
Now put all the given values in this formula, we get the angular acceleration.
![\alpha=((\gamma)-(\gamma-\beta t^2))/(t)](https://img.qammunity.org/2019/formulas/chemistry/high-school/hcqnrg0135i8bslezbizw12nk0k65haj25.png)
![\alpha=-\beta t=-0.810\text{ t}](https://img.qammunity.org/2019/formulas/chemistry/high-school/pfir0shafutdn8ftbrhf19zflehgom469w.png)
Therefore, the angular acceleration is
![\alpha=-\beta t=-0.810\text{ t}](https://img.qammunity.org/2019/formulas/chemistry/high-school/pfir0shafutdn8ftbrhf19zflehgom469w.png)