The pulley shown in the figure is made using | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the radius of the rim be $r$. The moment of inertia $I$ of the pulley consists of the rim and two rods.$I = I_{	ext{rim}} + 2 	imes I_{	ext

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the radius of the rim be $r$. The moment of inertia $I$ of the pulley consists of the rim and two rods.$I = I_{	ext{rim}} + 2 	imes I_{	ext{rod}}$$I = Mr^2 + 2 	imes \left( \frac{M(2r)^2}{12} ight) = Mr^2 + 2 	imes \left( \frac{4Mr^2}{12} ight) = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$.Let $a$ be the acceleration of the blocks and $T_1, T_2$ be the tensions in the string.The equations of motion are:$Mg - T_2 = Ma$ ... $(1)$$T_1 - mg = ma$ ... $(2)$$(T_2 - T_1)r = I \alpha = I \left( \frac{a}{r} ight) \implies T_2 - T_1 = \frac{I}{r^2} a = \frac{5}{3}Ma$ ... $(3)$Adding $(1)$, $(2)$, and $(3)$:$(M - m)g = (M + m + \frac{5}{3}M)a$$(M - m)g = (\frac{8}{3}M + m)a$$a = \frac{(M - m)g}{\frac{8}{3}M + m}$.

Similar Questions

What is a rigid body?

The direction of the angular velocity vector is along which of the following?

As shown in the figure,a mass $m = 500 \ g$ hangs from the rim of a wheel of radius $r = 20 \ cm$. When released from rest,the mass falls $2.0 \ m$ in $8 \ s$. Then the moment of inertia of the wheel is .......... $kg \cdot m^2$. $(g = 10 \ m/s^2)$

$A$ binary star system consists of two stars,one of which has double the mass of the other. The stars rotate about their common centre of mass:

$A$ disc rotating about its axis from rest,acquires the angular speed $100 \, rev/s$ in $4 \, s$. The angle rotated by it during these four seconds (in radian) is ...... $\pi$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module