Question:

A rope thrown over a pulley has a ladder with a man of mass m on one of its ends and a counter balancing mass M on its other end. The man climbs with a velocity VrV_r relative to ladder . Ignoring the masses of the pulley and the rope as well as the friction on the pulley axis, the velocity of the center of mass of this system is :

Updated On: Jul 6, 2022
  • mMVr\frac{m}{M}V_r
  • m2MVr\frac{m}{2M}V_r
  • MmVr\frac{M}{m}V_r
  • 2MmVr\frac{2M}{m}V_r
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The Correct Option is B

Solution and Explanation

The masses of load, ladder and man are M, M-m and m respectively. Their velocities are V(upward), -V and VrV_r -V respectively ∴\therefore Vcm=∑m1V1∑m1V_{cm} = \frac{\sum m_1 V_1}{\sum m_1} = M(ν)+(M−m)(−ν)+m(Vr−ν)2M\frac{M(\nu) + (M - m )(-\nu) +m(V_r - \nu)}{2M} = m2MVr\frac{m}{2M } V_r
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Concepts Used:

Center of Mass

The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.

The formula for the Centre of Mass:

Center of Gravity

The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.