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 diagonals,
I = $\frac{m a^{2}}{12}$
Therefore, we have:
Torque produced, $τ = I α$

$\Rightarrow τ = (\frac{m a^{2}}{12}) \times α = \frac{120 \times 10^{- 3} \times 5^{2} \times 10^{- 4}}{12} \times 0.2 = 10 \times 25 \times 10^{- 4} \times 0.2 = 0.5 \times 10^{- 5} N - m$

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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 diagonals, I = ma212 Therefore, we have: Torque produced, τ=Iα ⇒τ=ma212×α =120×10-3×52×10-412 ×0.2 =10×25×10-4× 0.2 =0.5×10-5 N-m