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Question Number 22135 by Tinkutara last updated on 11/Oct/17

A mass m hangs with the help of a  string wrapped around a pulley on a  frictionless bearing. The pulley has  mass m and radius R. Assuming pulley  to be a perfect uniform circular disc, the  acceleration of the mass m, if the  string does not slip on the pulley, is

$$\mathrm{A}\:\mathrm{mass}\:{m}\:\mathrm{hangs}\:\mathrm{with}\:\mathrm{the}\:\mathrm{help}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{string}\:\mathrm{wrapped}\:\mathrm{around}\:\mathrm{a}\:\mathrm{pulley}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{frictionless}\:\mathrm{bearing}.\:\mathrm{The}\:\mathrm{pulley}\:\mathrm{has} \\ $$$$\mathrm{mass}\:{m}\:\mathrm{and}\:\mathrm{radius}\:{R}.\:\mathrm{Assuming}\:\mathrm{pulley} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{uniform}\:\mathrm{circular}\:\mathrm{disc},\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mass}\:{m},\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{string}\:\mathrm{does}\:\mathrm{not}\:\mathrm{slip}\:\mathrm{on}\:\mathrm{the}\:\mathrm{pulley},\:\mathrm{is} \\ $$

Answered by ajfour last updated on 11/Oct/17

mg−T=ma     ...(i)  TR=(((mR^2 )/2))α  ;   αR=a  so     T=ma/2    ...(ii)  adding (i) and (ii)  mg=((3ma)/2)       ⇒    a=((2g)/3) .

$${mg}−{T}={ma}\:\:\:\:\:...\left({i}\right) \\ $$$${TR}=\left(\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\right)\alpha\:\:;\:\:\:\alpha{R}={a} \\ $$$${so}\:\:\:\:\:{T}={ma}/\mathrm{2}\:\:\:\:...\left({ii}\right) \\ $$$${adding}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${mg}=\frac{\mathrm{3}{ma}}{\mathrm{2}}\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\boldsymbol{{a}}=\frac{\mathrm{2}\boldsymbol{{g}}}{\mathrm{3}}\:. \\ $$

Commented by Tinkutara last updated on 11/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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