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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

A system of two identical, uniform discs with identical circular...

A system of two identical, uniform discs with identical circular cavities, is shown in the figure. Different relevant coordinates are given in the figure. The coordinates of the centre of mass of the system are

& A system of two identical, uniform discs with identical circular cavities, is shown in the figure. Different relevant

Options

  1. A(3R2,5R4)\left(\dfrac{3\text{R}}{2}\text{,}\dfrac{5\text{R}}{4}\right)
  2. B(19R6,5R2)\left(\dfrac{19\text{R}}{6}\text{,}\dfrac{5\text{R}}{2}\right)
  3. C(R2,R4)\left(\dfrac{R}{2}\text{,}\dfrac{R}{4}\right)
  4. D(20R6,5R2)\left(\dfrac{20\text{R}}{6}\text{,}\dfrac{5\text{R}}{2}\right)

Correct answer

B. (19R6,5R2)\left(\dfrac{19\text{R}}{6}\text{,}\dfrac{5\text{R}}{2}\right)

Step-by-step solution

Mass of the removed portion

=M\piR2×π(R2)2= \dfrac{M}{\piR^{2}} \times \pi\left(\dfrac{R}{2}\right)^{2}

=M4= \dfrac{M}{4}

xcm=M⋅0−M4×R+M×7R−M4×8RM−M4+M−M4{x}_{cm} = \dfrac{\text{M} \cdot 0 - \dfrac{M}{4} \times \text{R} + \text{M} \times 7\text{R} - \dfrac{M}{4} \times 8\text{R}}{\text{M} - \dfrac{M}{4} + \text{M} - \dfrac{M}{4}}

=19R6= \dfrac{19\text{R}}{6}

ycm=M⋅0−M40+M5R−M45RM−M4+M−M4{y}_{cm} = \dfrac{\text{M} \cdot 0 - \dfrac{M}{4}0 + \text{M}5\text{R} - \dfrac{M}{4}5\text{R}}{\text{M} - \dfrac{M}{4} + \text{M} - \dfrac{M}{4}}

=5R2= \dfrac{5\text{R}}{2}

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