Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with the same angular acceleration.

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Solution

Moment of inertia of a square plate about its diagonal is $\frac{m a^{2}}{12}$

(m = mass of the square plate.)

a = edges of the square

Therefore, torque produed

$\frac{m a^{2}}{12}$

${\frac{(120 \times 10^{- 3} \times 5.2 \times 10^{- 4})}{12 + 0.2}}$

$= 10 \times 25 \times 10^{- 4} \times 0.2$

$= 0.5 \times 10^{- 5} N - m$

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Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with the same angular acceleration.

Moment of inertia of a square plate about its diagonal is ma212 (m = mass of the square plate.) a = edges of the square Therefore, torque produed ma212 {(120×10−3×5.2×10−4)12+0.2} =10×25×10−4×0.2 =0.5×10−5N−m