Answer:
Explanation:
Given
radius r=2.96 mm
Tension T=2.4 N
time taken=0.74 s
Let [tex]\alpha [/tex]be the angular acceleration
[tex]2 T\times r=I\times \alpha [/tex]
[tex]2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha [/tex]
[tex]\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}[/tex]
[tex]\alpha =\frac{3.24\times 10^3}{m} rad/s^2[/tex]
[tex]\omega =\omega _0+\alpha \cdot t[/tex]
[tex]\omega =0+\frac{3.24\times 10^3}{m}\times 0.74[/tex]
[tex]\omega =\frac{2.4\times 10^3}{m} rad/s[/tex]
Angular momentum
[tex]L=I\omega [/tex]
[tex]L=0.5\times mr^2\times \omega [/tex]
[tex]L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}[/tex]
[tex]L=0.01051 kg-m^2/s[/tex]
