Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

As shown in the figure, a chain of mass $m$ is placed on a smooth quarter circular portion of radius $R$ . End $A$ is tied with a wedge while the remaining chain is free, then the minimum work required to be done by the external agent to make the chain horizontal keeping point $A$ fixed, is

Options:

Answer:

Solution:

Center of mass of the chain is at $\frac{2 R}{π}$ from horizontal

As shown in figure .......

Height gain by center of mass is $\frac{2 R}{π}$

$Wg = - m g (\frac{2 R}{π})$

$W_{ext.} = - Wg = (m g \frac{2 R}{π})$

Alternate solution

$U_{i} = \int - (dm) gy$

$= g \int R cos θ (\frac{m}{\frac{π}{2}}) d θ$

$= - \frac{2 Rg m}{π} \int_{0}^{\frac{π}{2}} cos θ d θ$

$= - \frac{2 Rg m}{π} [sin θ]_{0}^{π / 2}$

$= - \frac{2 m gR}{π}$

$U_{f} = 0 \Rightarrow Δ U = w_{ext} = \frac{2 m gR}{π}$

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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