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Standard XII

Physics

Ropes

The pulley sh...

Question

The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.5 kg- $m^{2}$ about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.

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Solution

$m_{1} g s i n θ - T_{1} = m_{1} a (1)$

$T_{1} - T_{2} = I \frac{a}{r^{2}} (2)$

$T_{2} - m_{2} g s i n θ = m_{2} a (3)$

$Adding the equations (1) and (3), we will get,$

$m_{1} g s i n θ + (T_{2} - T_{1}) - m_{2} g s i n θ$

$= (m_{1} + m_{2}) a$

$\Rightarrow (m_{1} - m_{2}) g s i n θ$

$= (m_{1} + m_{2} + \frac{I}{r^{2}}) a$

$\Rightarrow a = \frac{(m 1 - m 2) g s i n θ}{(m_{1} + m_{2} + \frac{I}{r_{2}})}$

$= \frac{(4 - 2) \times 10 \times (1 \sqrt{2})}{{(4 + 2) + (0.5 / 0.01)}}$

$= \frac{(2 \times 10 \times \frac{1}{\sqrt{2}})}{(6 + 50)}$

$= 0.248 = 0.25 m / s^{2}$

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The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.5 kg-m2 about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.

m1g sin θ−T1=m1a (1) T1−T2=Iar2 (2) T2−m2g sin θ=m2a (3) Adding the equations(1)and(3),we will get, m1g sin θ+(T2−T1)−m2g sin θ =(m1+m2)a ⇒(m1−m2)g sin θ =(m1+m2+Ir2)a ⇒a=(m1−m2)g sin θ(m1+m2+Ir2) =(4−2)×10×(1√2){(4+2)+(0.5/0.01)} =(2×10×1√2)(6+50) =0.248=0.25 m/s2

The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.5 kg- $m^{2}$ about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.