The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass M and m are attached to two ends of a light string passing over the pulley, which is hinged

Question

The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitude of the acceleration experienced by the blocks is ______ (assume no slipping of string on pulley). [2026]

Smart Achievers · Accepted Answer

(1)

$M g - T_{2} = M a . . . (1)$

$T_{1} - m g = m a . . . (2)$

$(T_{2} - T_{1}) r = I \frac{a}{r} . . . (3)$

$(1) + (2) + (3)$

$(M - m) g = (M + m + \frac{I}{r^{2}}) a$

$Here I = M r^{2} + \frac{M \times {(2 r)}^{2}}{12} \times 2$

$= (1 + \frac{2}{3}) M r^{2}$

$= \frac{5}{3} M r^{2}$

$(M - m) g = [M + m + \frac{5 M}{3}] a$

$a = \frac{(M - m) g}{[M + m + \frac{5 M}{3}]}$

Solution

1 $\frac{(M - m) g}{[(\frac{8}{3}) M + m]}$

2 $\frac{(M - m) g}{2 M + m}$

3 $\frac{(M - m) g}{M + m}$

4 $\frac{(M - m) g}{[(\frac{13}{6}) M + m]}$

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Solution

1 (M-m)g[(83) M+m]

2 (M-m)g2M+m

3 (M-m)gM+m

4 (M-m)g[(136)M+m]

Ans. (1) Mg-T2=Ma ...(1) T1-mg=ma ...(2) (T2-T1)r=Iar ...(3) (1)+(2)+(3) (M-m)g=(M+m+Ir2)a Here I=Mr2+M×(2r)212×2 =(1+23)Mr2 =53Mr2 (M-m)g=[M+m+5M3]a a=(M-m)g[M+m+5M3]

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1 $\frac{(M - m) g}{[(\frac{8}{3}) M + m]}$

2 $\frac{(M - m) g}{2 M + m}$

3 $\frac{(M - m) g}{M + m}$

4 $\frac{(M - m) g}{[(\frac{13}{6}) M + m]}$