Two bodies $m_{1}$ and $m_{2}$ are attached to the two ends of a string figure. The string passes over a pulley of mass $M$ and radius $R$ . If $m_{1} > m_{2}$ , then the acceleration of the system is

\frac{m_{1} - m_{2} + m) g}{m_{1} + m_{2} + m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{{(m}_{1} + m_{2}) g}{m_{1} - m_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{{(m}_{1} - m_{2}) g}{m_{1} - m_{2} + m / 2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{{(m}_{1} - m_{2}) g}{m_{1} - m_{2} + m / 2}$
Force balance equations for two weights can be written as:
$T_{1} = m_{1} g - m_{1} a$ ---------(i)
and $T_{2} = m_{2} a + m_{2} g$ ----------(ii)
(ii)-(i), we get:
$T_{1} - T_{2} = (m_{1} - m_{2}) g - (m_{1} + m_{2}) a$ ------------(iii)
As torque, $τ = I α$ , we have:
$(T_{1} - T_{2}) R = (\frac{1}{2} M R^{2}) . \frac{a}{R}$
or, $T_{1} - T_{2} = \frac{M a}{2}$
Substituting this in equation (iii), we get:
$\frac{M a}{2} = (m_{1} - m_{2}) g - (m_{1} + m_{2}) a$
or, $a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2} + (m / 2)}$

Suggest Corrections

Similar questions

Q. A circular pulley of mass

^{'} M^{'}

and radius

^{'} R^{'}

, is hinged at the centre. A long string is wound over this disc and two bodies of mass

m_{1}

and

m_{2}

are attached at it's free ends. Now the bodies are released. Find the accceleration of each body.

Q. A circular pulley of mass

^{'} M^{'}

and radius

^{'} R^{'}

, is hinged at the centre. A long string is wound over this disc and two bodies of mass

m_{1}

and

m_{2}

are attached at its free ends. Now the bodies are released. Find the acceleration of each body.

Two bodies of masses m₁ and m₂ are connected by a light, inextensible string, which passes over a frictionless pulley. If the pulley is moving upward with uniform acceleration g, then the tension in the string is

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:

Q. The two bodies of masses

m_{1}

and

m_{2} (m_{1} > m_{2})

respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are initially at rest and released. The acceleration of the centre of mass is

Join BYJU'S Learning Program

Two bodies m1 and m2 are attached to the two ends of a string figure. The string passes over a pulley of mass M and radius R. If m1>m2, then the acceleration of the system is

Two bodies $m_{1}$ and $m_{2}$ are attached to the two ends of a string figure. The string passes over a pulley of mass $M$ and radius $R$ . If $m_{1} > m_{2}$ , then the acceleration of the system is