A rectangular plate of mass 20 kg is suspend | vedclass

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(B) The situation after removing pin $B$ is shown in the figure. Let the length and width be $l = 0.2 \ m$ and $b = 0.15 \ m$ respectively.When pin $B

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(B) The situation after removing pin $B$ is shown in the figure. Let the length and width be $l = 0.2 \ m$ and $b = 0.15 \ m$ respectively.When pin $B$ is removed,the plate starts rotating about point $A$.The torque $	au$ due to gravity acting at the center of mass is given by:$	au = mg \left( \frac{l}{2} ight)$Using the relation $	au = I_A \alpha$,where $I_A$ is the moment of inertia about point $A$:$I_A \alpha = mg \left( \frac{l}{2} ight) \quad \dots(i)$The moment of inertia of a rectangular plate about its center of mass is $I_{CM} = \frac{M}{12}(l^2 + b^2)$.Using the parallel axis theorem,$I_A = I_{CM} + Md^2$,where $d$ is the distance from the center of mass to point $A$:$d^2 = \left( \frac{l}{2} ight)^2 + \left( \frac{b}{2} ight)^2 = \frac{0.2^2}{4} + \frac{0.15^2}{4} = \frac{0.04 + 0.0225}{4} = \frac{0.0625}{4} = 0.015625 \ m^2$$I_A = \frac{20}{12}(0.2^2 + 0.15^2) + 20(0.015625) = \frac{20}{12}(0.04 + 0.0225) + 0.3125 = \frac{20}{12}(0.0625) + 0.3125 = 0.104167 + 0.3125 = 0.416667 \ kg \cdot m^2$Now,substitute into equation $(i)$:$0.416667 \alpha = 20 	imes 9.8 	imes \left( \frac{0.2}{2} ight)$$0.416667 \alpha = 196 	imes 0.1 = 19.6$$\alpha = \frac{19.6}{0.416667} \approx 47.04 \ rad/s^2$. Given the options,the intended calculation uses $g = 10 \ m/s^2$:$0.416667 \alpha = 20 	imes 10 	imes 0.1 = 20$$\alpha = \frac{20}{0.416667} = 48 \ rad/s^2$.

$A$ rectangular plate of mass $20 \ kg$ is suspended from points $A$ and $B$ as shown in the figure. If the pin at $B$ is suddenly removed,find the angular acceleration of the plate in $rad/s^2$.

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