
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
${v_r}$ 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:
$
{\text{A}}{\text{. }}\dfrac{m}{M}{v_r} \\
{\text{B}}{\text{. }}\dfrac{m}{{2M}}{v_r} \\
{\text{C}}{\text{. }}\dfrac{M}{m}{v_r} \\
{\text{D}}{\text{. }}\dfrac{{2M}}{m}{v_r} \\
$
Answer
596.1k+ views
Hint: In this question first we have to find out the mass of the ladder then we have to apply the concept of relative velocity to find out the velocity of each mass with respect to ground and finally apply the formula for finding the center of mass of the given system.
Formulas used:
Relative velocity of body A with respect to body B = velocity of the body A – velocity of the body B
${V_{cm}} = \dfrac{{{m_1} \times {v_1} + {m_2} \times {v_2} + ......{m_n} \times {v_n}}}{{{m_1} + {m_2} + ......{m_n}}}$
Complete Step-by-Step solution:
Given:- Mass of man= m
Counterbalancing mass on other side of pulley= M
Then, mass of ladder=M-m
If a man moves up, the ladder moves down and counterbalancing mass moves up.
Velocity of ladder= V
Velocity of man relative to ladder=${v_r}$
Then, velocity of man relative to ground= ${v_r}$- V
${V_{cm}} = \dfrac{{{m_1} \times {v_1} + {m_2} \times {v_2} + ......{m_n} \times {v_n}}}{{{m_1} + {m_2} + ......{m_n}}}$
where ${m_1},{m_2}....$ are masses of system and ${v_1},{v_2}....$ are their respective velocities. Taking upside direction is positive and downward is negative.
Then,
$ \Rightarrow {V_{cm}} = \dfrac{{MV + m({v_r} - V) - (M - m)V}}{{M + m + M - m}}$
On further solving the above equation we get
$ \Rightarrow {V_{cm}} = \dfrac{{m{v_r}}}{{2M}}$
Hence, option B. is correct.
Note:- Whenever you get this type of question the key concept to solve is to learn the concept of relative velocity which states that the relative velocity of a body A w.r.t. to another body B is the velocity that body A would appear to have to an observer situated on body B moving along with it and the concept of center of mass.
Formulas used:
Relative velocity of body A with respect to body B = velocity of the body A – velocity of the body B
${V_{cm}} = \dfrac{{{m_1} \times {v_1} + {m_2} \times {v_2} + ......{m_n} \times {v_n}}}{{{m_1} + {m_2} + ......{m_n}}}$
Complete Step-by-Step solution:
Given:- Mass of man= m
Counterbalancing mass on other side of pulley= M
Then, mass of ladder=M-m
If a man moves up, the ladder moves down and counterbalancing mass moves up.
Velocity of ladder= V
Velocity of man relative to ladder=${v_r}$
Then, velocity of man relative to ground= ${v_r}$- V
${V_{cm}} = \dfrac{{{m_1} \times {v_1} + {m_2} \times {v_2} + ......{m_n} \times {v_n}}}{{{m_1} + {m_2} + ......{m_n}}}$
where ${m_1},{m_2}....$ are masses of system and ${v_1},{v_2}....$ are their respective velocities. Taking upside direction is positive and downward is negative.
Then,
$ \Rightarrow {V_{cm}} = \dfrac{{MV + m({v_r} - V) - (M - m)V}}{{M + m + M - m}}$
On further solving the above equation we get
$ \Rightarrow {V_{cm}} = \dfrac{{m{v_r}}}{{2M}}$
Hence, option B. is correct.
Note:- Whenever you get this type of question the key concept to solve is to learn the concept of relative velocity which states that the relative velocity of a body A w.r.t. to another body B is the velocity that body A would appear to have to an observer situated on body B moving along with it and the concept of center of mass.
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