
A uniform metal disc of radius R is taken and out of it, a disc of diameter R is cut-off from the end. The centre of mass of the remaining part will be:
A: R/4 from the centre
B: R/3 from the centre
C: R/5 from the centre
D: R/6 from the centre
Answer
481.2k+ views
Hint: When we consider the given conditions, we can conclude from them that the density of the disc will be uniform. Hence we can geometrically compare the dimensions in both cases and then calculate using those values to generate the desired result.
Complete step by step answer:
We can use the formula for centre of mass as: x coordinate of the centre of mass of a system
\[{{x}_{cm}}=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{x}_{i}}}}{\sum{{{m}_{i}}}}\] where m is the mass and x is the position.
Similarly, y coordinate of the centre of mass of the system is
\[{{y}_{cm}}=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{y}_{i}}}}{\sum{{{m}_{i}}}}\]
Hence, the coordinates of the centre of mass is given by: $({{x}_{cm}},{{y}_{cm}})$
We are given a uniform metal disc of radius R and another disc that is cut out from the bigger one. The radius of the smaller disc is R/2. We have to find the centre of mass of the remaining disc.
We can geometrically approach this problem.
Area of bigger disc $A=\pi {{R}^{2}}$
Area of the cut off disc $a=\pi {{(\dfrac{R}{2})}^{2}}=\dfrac{\pi {{R}^{2}}}{4}$
We know that
\[{{x}_{cm}}=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{x}_{i}}}}{\sum{{{m}_{i}}}}\]
Here,
\[{{x}_{cm}}=\dfrac{{{A}_{1}}\times 0-{{A}_{2}}{{X}_{2}}}{{{A}_{1}}-{{A}_{2}}}\]
\[{{x}_{cm}}=\dfrac{-\dfrac{\pi {{R}^{2}}}{4}\times \dfrac{R}{2}}{\pi {{R}^{2}}-\dfrac{\pi {{R}^{2}}}{4}}=-\dfrac{R}{6}\]
Hence, the centre of mass of the remaining portion of the disc lies at a distance R/6 from the centre.
Therefore, option D is the correct answer among the given options.
Note: The centre of mass has various applications in space studies as it is used as a reference point for calculations in mechanics which involve the masses from space, like the linear momentum and the angular momentum of planetary bodies and the dynamics of rigid bodies.
Complete step by step answer:
We can use the formula for centre of mass as: x coordinate of the centre of mass of a system
\[{{x}_{cm}}=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{x}_{i}}}}{\sum{{{m}_{i}}}}\] where m is the mass and x is the position.
Similarly, y coordinate of the centre of mass of the system is
\[{{y}_{cm}}=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{y}_{i}}}}{\sum{{{m}_{i}}}}\]
Hence, the coordinates of the centre of mass is given by: $({{x}_{cm}},{{y}_{cm}})$
We are given a uniform metal disc of radius R and another disc that is cut out from the bigger one. The radius of the smaller disc is R/2. We have to find the centre of mass of the remaining disc.
We can geometrically approach this problem.
Area of bigger disc $A=\pi {{R}^{2}}$
Area of the cut off disc $a=\pi {{(\dfrac{R}{2})}^{2}}=\dfrac{\pi {{R}^{2}}}{4}$
We know that
\[{{x}_{cm}}=\dfrac{\sum\limits_{i=1}^{n}{{{m}_{i}}{{x}_{i}}}}{\sum{{{m}_{i}}}}\]
Here,
\[{{x}_{cm}}=\dfrac{{{A}_{1}}\times 0-{{A}_{2}}{{X}_{2}}}{{{A}_{1}}-{{A}_{2}}}\]
\[{{x}_{cm}}=\dfrac{-\dfrac{\pi {{R}^{2}}}{4}\times \dfrac{R}{2}}{\pi {{R}^{2}}-\dfrac{\pi {{R}^{2}}}{4}}=-\dfrac{R}{6}\]
Hence, the centre of mass of the remaining portion of the disc lies at a distance R/6 from the centre.
Therefore, option D is the correct answer among the given options.
Note: The centre of mass has various applications in space studies as it is used as a reference point for calculations in mechanics which involve the masses from space, like the linear momentum and the angular momentum of planetary bodies and the dynamics of rigid bodies.
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