Rotational Mechanics — Physics STD 12 Science — Question

m₁ = 4kg, m₂ = 2kg

Frictional co-efficient between 2kg block and surface = 0.5

R = 10cm = 0.1m

I = 0.5kg-m²

$\text{m}_1\text{g}\sin\theta-\text{T}_1=\text{m}_1\text{a}\ \dots(1)$

$\text{T}_2-(\text{m}_2\text{g}\sin\theta+\mu\text{m}_2\text{g}\cos\theta)=\text{m}_2\text{a}\ \dots(2)$

$(\text{T}_1-\text{T}_2)=\frac{\text{la}}{\text{r}^2}$

Adding equation (1) and (2) we will get

$\text{m}_1\text{g}\sin\theta-(\text{m}_2\text{g}\sin\theta+\mu\text{m}_2\text{g}\cos\theta)+(\text{T}_1+\text{T}_2)=\text{m}_1\text{a}+\text{m}_2\text{a}$

$\Rightarrow4\times9.8\times\Big(\frac{1}{\sqrt2}\Big)-\Big\{\Big(2\times9.8\times\Big(\frac{1}{\sqrt2}\Big)+0.5\times2\times9.8\times\Big(\frac{1}{\sqrt2}\Big)\Big\}$

$=\Big(4+2+\frac{0.5}{0.01}\Big)\text{a}$

$\Rightarrow27.80-(13.90+6.95)=65\text{a}$

$\Rightarrow\text{a}=0.125\text{ms}^{-2}$