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Exercises · 6.15

Q.From a uniform disk of radius RR, a circular hole of radius R/2R/2 is cut out. The centre of the hole is at R/2R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

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Treat the hole as negative mass at distance R/2R/2 from the disk's center; the center of gravity shifts to −R/6\boxed{-R/6} from the original center, along the line joining the two centers (away from the hole).

The center of mass of a composite body can be found by treating it as a superposition of simpler shapes. When material is removed, we imagine adding a "negative mass" at the location of the hole. This transforms a subtraction problem into an addition problem, letting us use the standard center-of-mass formula.

The original disk is uniform, so its center of mass sits at its geometric center. The hole is also circular and uniform, so its center of mass would be at its geometric center, located at distance R/2R/2 from the disk's center. The resulting body is the original disk minus the hole.


Step-by-step calculation:

  1. Set up coordinates. Place the origin at the center of the original disk. Let the center of the hole lie along the positive xx-axis at x=R/2x = R/2.

  2. Assign masses. Let the surface mass density be σ\sigma (mass per unit area). The original disk has area πR2\pi R^2, so its mass is:

Mdisk=σπR2M_{\text{disk}} = \sigma \pi R^2

The hole has radius R/2R/2, so its area is π(R/2)2=πR2/4\pi (R/2)^2 = \pi R^2/4, and its "negative mass" is:

Mhole=−σπR24M_{\text{hole}} = -\sigma \frac{\pi R^2}{4}

The mass of the resulting body is:

Mfinal=σπR2−σπR24=3σπR24M_{\text{final}} = \sigma \pi R^2 - \sigma \frac{\pi R^2}{4} = \frac{3\sigma \pi R^2}{4}

  1. Locate the individual centers of mass. The disk's center of mass is at the origin: xdisk=0x_{\text{disk}} = 0. The hole's center is at xhole=R/2x_{\text{hole}} = R/2.

  2. Apply the center-of-mass formula. For a system of masses mim_i at positions xix_i, the center of mass is:

xcm=∑mixi∑mix_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}

Substituting our values:

xcm=Mdisk⋅0+Mhole⋅(R/2)Mfinalx_{\text{cm}} = \frac{M_{\text{disk}} \cdot 0 + M_{\text{hole}} \cdot (R/2)}{M_{\text{final}}}

xcm=0+(−σπR24)⋅R23σπR24x_{\text{cm}} = \frac{0 + \left(-\sigma \frac{\pi R^2}{4}\right) \cdot \frac{R}{2}}{\frac{3\sigma \pi R^2}{4}} …

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