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Rotational Mechanics — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsRotational Mechanics5 Marks
Question
Solve the previous problem if the friction coefficient between the 2.0kg block and the plane below it is 0.5 and the plane below the 4.0kg block is frictionless.
✓

Answer


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

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

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

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

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

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

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