Answer:
[tex]9.16rad/s^2[/tex]
Explanation:
We are given that
Mass,[tex]m_1=3.3 kg[/tex]
Radius,r=0.8 m
[tex]m_2=4.9 kg[/tex]
Height,h=2.9 m
We have to find the angular acceleration of the cylinder.
According to question
[tex]4.9g-T=4.9a[/tex]
[tex]Tr=I\alpha[/tex]
Where
[tex]\alpha=\frac{a}{r}[/tex]
[tex]Tr=\frac{1}{2}m_1ra[/tex]
[tex]T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a[/tex]
Substitute the value
[tex]4.9g-\frac{1}{2}(3.3a)=4.9a[/tex]
[tex]4.9\times 9.8=4.9a+\frac{3.3a}{2}[/tex]
Where [tex]g=9.8 m/s^2[/tex]
[tex]48.02=a(4.9+1.65)=6.55a[/tex]
[tex]a=\frac{48.02}{6.55}=7.33m/s^2[/tex]
Angular acceleration,[tex]\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2[/tex]
