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  • A massless string connects two pulley of masses ' ' and '<img alt="1 \mat

A massless string connects two pulley of masses ' 2 \mathrm{~kg}' and '1 \mathrm{~kg}' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2


Option: 2

\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2


Option: 3

\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2


Option: 4

\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2


Answers (2)

best_answer

For the translatory motion of the smaller pulley

m_2 g-T=m_2 a~~~~~~~~...(i)

For rotational motion.

\begin{aligned}T\left(r\right)&=\frac{1}{2} m_2 r_2^2 \alpha_2 \\ T&=\frac{m_2 r_2 \alpha_2}{2}~~~~~~...(ii) \end{aligned}

For the bigger pulley

\begin{aligned} T\left(r_1\right) & =\frac{1}{2} m_1 r_1^2 \alpha_1 \\ T & =\frac{m_1 r_1 \alpha_1}{2}~~~~~~\text { ...(iii) } \end{aligned}

\begin{aligned} r_1 \alpha_1+r_2 \alpha_2&=a \\ \frac{2 T}{m_1}+\frac{2 T}{m_2}&=a \\ 2 T\left[\frac{1}{m_1}+\frac{1}{m_2}\right]&=a \\ 2 T\left[\frac{m_1+m_2}{m_1 m_2}\right]&=a \end{aligned}

From equation (i)
\begin{aligned} m_2 g-T & =m_2 a \\ m_2 g & =T+m_2 a \\ m_2 g & =\left(\frac{m_1 m / L}{m_1+m_2}\right) \frac{a}{2}+m_2 a \\ g & =a\left[\frac{m_1}{2\left(m_1+m_2\right)}+1\right] \\ a & =\frac{2\left(m_1+m_2\right)}{\left(3 m_1+2 m_2\right)} g \\ a & =\frac{2(2+1)}{(3 \times 2+2 \times 1)} \times 10 \\ a & =\frac{x \times 3 \times 10}{84}=\frac{30}{4}=7.5 \mathrm{~m} / \mathrm{s}^2 \end{aligned}

Posted by

manish painkra

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Not understanding sir 

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Raju vittal nandi

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