Answer:
Here, m = 10 kg; R = 15 cm = 0.15 m; T = 1.5s, \[\left[
\text{T}
\right]\text{=}\frac{\text{1}}{\text{4}}\text{m}{{\text{a}}^{\text{2}}}{{\text{
}\!\!\omega\!\!\text{ }}^{\text{2}}}\]=?
Moment of inertia of disc,
\[{{\text{E}}_{{{\text{K}}_{\text{av}}}}}\text{=}{{\text{E}}_{{{\text{P}}_{\text{av}}}}}\]
Now, \[1\cdot 5\]
So, \[\alpha \]
\[\alpha \theta \]
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