Q. The moment of inertia of the pulley system as shown in the figure- $5.103$ is $4\, kgm ^{2}$ . The radii of bigger and smaller pulleys $2\, m$ and $1 \,m$ respectively. The angular acceleration of the pulley system is: (take $g=10\, m / s ^{2}$ )

System of Particles and Rotational Motion Report Error

A

$2.1 \,rad / s ^{2}$

B

$4.2 \,rad / s ^{2}$

C

$1.2\, rad / s ^{2}$

D

$0.6\, rad / s ^{2}$

Solution:

From figure shown

$50-T_{1}=5(2 \alpha)\,\,\,...(1)$
$T_{2}-40=4(\alpha)\,\,\,...(2)$
$T_{1}(2)-T_{2}(1)=4 \alpha\,\,\,...(3)$
From (1), (2) and (3)
$2(50-10 \alpha)-(40+4 \alpha) =4 \alpha$
$\Rightarrow 100-20 \alpha-40-4 \alpha =4 \alpha$
$\Rightarrow 28 \alpha =60$
$\Rightarrow \alpha =\frac{60}{28}=2.1 \,m / s ^{2}$

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Q. The moment of inertia of the pulley system as shown in the figure- 5.1035.103 is 4kgm24\, kgm ^{2}. The radii of bigger and smaller pulleys 2m2\, m and 1m1 \,m respectively. The angular acceleration of the pulley system is: (take g=10m/s2g=10\, m / s ^{2} )

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