Question

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

Answer 1

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Answer 2

The torque on the square plate is $5 \times 10^{-6}$ N.m.

Explanation:

Step 1:

Given values in the question :

           α  is Angular acceleration

           M is the mass

                      α = 0.20

                      M = 120 g  

Converting mass from gram to kilogram  :

                     $M=\frac{120}{1000}=0.12 \mathrm{kg}$  

                    Edge a = 5.0 cm

Converting from centimeter to meter  :

                   $a=\frac{5}{100}=0.05 \mathrm{m}$

Step 2:

Moment of plate inertia when the line goes through the center and parallel to the edge,

                       $\mathrm{I}=\frac{\mathrm{ma}^{2}}{12}$

Torque now (plate)

Step 3:

               T = Iα

                  $=\frac{m a^{2}}{12} \times 0.20 \ (\alpha=0.20)$

                  $=\frac{0.12 \times 0.05^{2}}{12} \times 0.20$

                  $=\frac{0.12 \times 2.5 \times 10^{-3}}{12} \times 0.20$

                 $=\frac{3 \times 10^{-4}}{12} \times 0.20$

                 $=2.5 \times 10^{-5} \times 0.20$

                 $=5 \times 10^{-6} \mathrm{N} \cdot \mathrm{m}$

Thus, the torque on the square plate is, T $=5 \times 10^{-6} \mathrm{N} \cdot \mathrm{m}$.