
A circular plate of uniform thickness has a diameter $56cm$. A circular portion of diameter $42cm$is removed from one edge as shown in the figure. The centre of mass of the remaining portion from the centre of plate will be:
A) $5cm$
B) $7cm$
C) $9cm$
D) $11cm$
Answer
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Hint: Recall the concept of centre of mass. The centre of mass of a body is a point at which the whole mass of the particles of the system is said to be concentrated. When any object or body is broken into fragments or parts, then this concept is used to analyse the motion of the particles of the body.
Complete step by step solution:
Step I:
Let the mass of the uniform circular plate is ${M_1}$ before one portion is removed.
The diameter is $d = 56cm$
So, radius will be $r = \dfrac{{56}}{2} = 28cm$
Centre of mass will be at the origin. Therefore, ${X_1} = 0$
Let the density of the plate be $\rho $. The density will be the same throughout.
Step II:
It is known that $\rho = \dfrac{{Mass}}{{Area}}$
$Mass = \rho \times Area$
$M = \rho \times (\pi {r^2})$
${M_1} = \rho \times \pi \times {(28)^2}$
Step III:
When the circular portion is removed then
The diameter given is $d = 42cm$
Radius will be $r = \dfrac{{42}}{2} = 21cm$
Centre of mass will be ${X_2} = 28 - 21 = 7cm$
Since one part is removed, therefore, it will have a negative sign.
${M_2} = - \rho \times \pi \times {(21)^2}$
Step IV:
The new center of mass will be given by the formula
${X_{COM}} = \dfrac{{{M_1}{X_1} - {M_2}{X_2}}}{{{M_1} - {M_2}}}$
Where ${X_1}$is the centre of mass of original disc
${X_2}$is the centre of mass of removed portion of the disc
Substituting all the values and solving for centre of mass,
${X_{COM}} = \dfrac{{\rho \pi {{(28)}^2}(0) - \rho \pi {{(21)}^2}(7)}}{{\rho \pi [{{(28)}^2} - {{(21)}^2}]}}$
${X_{COM}} = \dfrac{{0 - 3087}}{{784 - 441}}$
${X_{COM}} = \dfrac{{ - 3087}}{{343}}$
${X_{COM}} = - 9cm$
Step V:
$\therefore $The centre of mass of the remaining portion from the centre of plate will be $ = 9cm$
Hence, Option C is the right answer.
Note: It is important to note that the terms centre of mass and centre of gravity are different and not to be confused. The centre of mass of an object does not depend on its gravitational field. But the centre of gravity depends on the gravitational field. Also the centre of gravity is the point at which the distribution of the weight of a body is equal in all directions.
Complete step by step solution:
Step I:
Let the mass of the uniform circular plate is ${M_1}$ before one portion is removed.
The diameter is $d = 56cm$
So, radius will be $r = \dfrac{{56}}{2} = 28cm$
Centre of mass will be at the origin. Therefore, ${X_1} = 0$
Let the density of the plate be $\rho $. The density will be the same throughout.
Step II:
It is known that $\rho = \dfrac{{Mass}}{{Area}}$
$Mass = \rho \times Area$
$M = \rho \times (\pi {r^2})$
${M_1} = \rho \times \pi \times {(28)^2}$
Step III:
When the circular portion is removed then
The diameter given is $d = 42cm$
Radius will be $r = \dfrac{{42}}{2} = 21cm$
Centre of mass will be ${X_2} = 28 - 21 = 7cm$
Since one part is removed, therefore, it will have a negative sign.
${M_2} = - \rho \times \pi \times {(21)^2}$
Step IV:
The new center of mass will be given by the formula
${X_{COM}} = \dfrac{{{M_1}{X_1} - {M_2}{X_2}}}{{{M_1} - {M_2}}}$
Where ${X_1}$is the centre of mass of original disc
${X_2}$is the centre of mass of removed portion of the disc
Substituting all the values and solving for centre of mass,
${X_{COM}} = \dfrac{{\rho \pi {{(28)}^2}(0) - \rho \pi {{(21)}^2}(7)}}{{\rho \pi [{{(28)}^2} - {{(21)}^2}]}}$
${X_{COM}} = \dfrac{{0 - 3087}}{{784 - 441}}$
${X_{COM}} = \dfrac{{ - 3087}}{{343}}$
${X_{COM}} = - 9cm$
Step V:
$\therefore $The centre of mass of the remaining portion from the centre of plate will be $ = 9cm$
Hence, Option C is the right answer.
Note: It is important to note that the terms centre of mass and centre of gravity are different and not to be confused. The centre of mass of an object does not depend on its gravitational field. But the centre of gravity depends on the gravitational field. Also the centre of gravity is the point at which the distribution of the weight of a body is equal in all directions.
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