First slide

Methods to locate centre of mass

Question

A disc (of radius r cm) of uniform thickness and uniform density $σ$ has a square hole with sides of length I = $\frac{r}{\sqrt{2}}$ cm. One corner of the hole is located at the centre of the disc and centre of the hole lies on y-axis as shown. Then the y-coordinate of position of centre of mass of disc with hole (in cm) is

Moderate

A

$- \frac{r}{2 (π - \frac{1}{4})}$
B

$- \frac{r}{4 (π - \frac{1}{4})}$
C

$- \frac{r}{4 (π - \frac{1}{2})}$
D

$- \frac{3 r}{4 (π - \frac{1}{4})}$

Solution

This disc can be assumed to be made of a complete uniform disc and a square plate with same negative mass density.
$Y_{cm} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} = \frac{({πr}^{2}) σ (0) + l^{2} (- σ) (\frac{r}{2})}{{πr}^{2} σ + l^{2} (- σ)}$
= $\frac{- l^{2} r}{2 ({πr}^{2} - l^{2})} = \frac{- \frac{r^{2}}{2}}{2 ({πr}^{2} - \frac{r^{2}}{2})} = \frac{- r}{4 (π - \frac{1}{2})}$

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