A uniform thin metal plate of mass 10 kg wit | vedclass

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Question

(A) Let the surface mass density be $\sigma$. The total area of the plate is $A = (3 	imes 2) - (1 	imes 1) = 5 	ext{ units}^2$. Given mass $M = 10

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Let the surface mass density be $\sigma$. The total area of the plate is $A = (3 	imes 2) - (1 	imes 1) = 5 	ext{ units}^2$. Given mass $M = 10 \ kg$,so $\sigma = \frac{10}{5} = 2 \ kg/	ext{unit}^2$.We can consider the plate as a large rectangle of $3 	imes 2$ (mass $M_{total} = 3 	imes 2 	imes 2 = 12 \ kg$) minus a small square of $1 	imes 1$ (mass $m_{cut} = 1 	imes 1 	imes 2 = 2 \ kg$).The center of mass of the large rectangle is at $(1.5, 1.0)$.The center of mass of the cut-out square is at $(1.5, 1.5)$.Let the center of mass of the remaining plate be $(x, y)$. Using the principle of moments:$M_{total} X_{CM} = M_{plate} x + m_{cut} x_{cut} \Rightarrow 12(1.5) = 10(x) + 2(1.5)$$18 = 10x + 3 \Rightarrow 10x = 15 \Rightarrow x = 1.5$.$M_{total} Y_{CM} = M_{plate} y + m_{cut} y_{cut} \Rightarrow 12(1.0) = 10(y) + 2(1.5)$$12 = 10y + 3 \Rightarrow 10y = 9 \Rightarrow y = 0.9$.The ratio $\frac{x}{y} = \frac{1.5}{0.9} = \frac{15}{9}$.Comparing with $\frac{n}{9}$,we get $n = 15$.

$A$ uniform thin metal plate of mass $10 \ kg$ with the dimensions shown in the figure is given. The ratio of the $x$ and $y$ coordinates of the center of mass of the plate is $\frac{n}{9}$. The value of $n$ is:

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