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question 23 of chapter 10 rotation and rolling...///////////////////////////////
A square plate of mass 120 gm and edge 5 cm rotates about one of the edge.Let take a small area of the square of width dx and length a which is at a distance x from the axis ofrotation.Therefore mass of that small aream/a2 × a dx (m = mass of the square ; a = side of the plate)I = a03a0(m/ a2 ) ax2dx (m/ a)(x / 3)]= ma2/3Therefore torque produced = I × = (ma2/3) × = {(120 × 10–3 × 52 × 10–4)/3} 0.2= 0.2 × 10–4 = 2 × 10–5 N-m.
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