Consider two masses with $m_{1} > m_{2}$ connected by a light inextensible string that passes over a pulley of radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance $2 h$ . When the two masses pass each other, the speed of the masses is proportional to.

\sqrt{\frac{m_{1} - m_{2}}{m_{1} + m_{2} + \frac{I}{R^{2}}}}

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

\sqrt{\frac{(m_{1} + m_{2}) (m_{1} - m_{2})}{m_{1} + m_{2} + \frac{I}{R^{2}}}}

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

\sqrt{\frac{m_{1} + m_{2} + \frac{I}{R^{2}}}{m_{1} - m_{2}}}

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

\sqrt{\frac{\frac{I}{R^{2}}}{m_{1} + m_{2}}}

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

Open in App

Solution

The correct option is A $\sqrt{\frac{m_{1} - m_{2}}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
The objects meet when each has travelled distance $h$

The lost $P E = (m_{1} - m_{2}) g h$

This equals the gain in $K E$ , so

$K E = (m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} I {(\frac{V}{R})}^{2}$

Assuming that $R$ was given in your diagram as the radius of the pulley.

$(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2} + I / R^{2}) v^{2} v = \sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + I / R^{2}}} v α \sqrt{\frac{m_{1} - m_{2}}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

The objects meet when each has travelled distance $h$

The lost $P E = (m_{1} - m_{2}) g h$

This equals the gain in $K E$ , so

$K E = (m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} I {(\frac{V}{R})}^{2}$

Assuming that $R$ was given in your diagram as the radius of the pulley.

$(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2} + I / R^{2}) v^{2} v = \sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + I / R^{2}}} v α \sqrt{\frac{m_{1} - m_{2}}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

Suggest Corrections

Similar questions

Q. Consider two objects with masses

m_{1}

and

m_{2}

(

m_{1} > m_{2}

) connected by a light string that passes over a pulley having a moment of inertia of

I

about its axis of rotation as shown in figure. The string does not slip on the pulley or stretch. The pulley turns without fiction. The two objects are released from rest separated by a vertical distance 2

h

. The translational speeds of the objects as they pass each other 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

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

Q. Two masses

m_{1}

and

m_{2}

are connected by a string passing over pulley. When set free,

m_{1}

moves downwards by

1.4 m

2 s

. The ratio between

m_{1}

and

m_{2}

Join BYJU'S Learning Program