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question 23 of chapter 10 rotation and rolling...///////////////////////////////

question 23 of chapter 10 rotation and rolling...///////////////////////////////

Grade:11

1 Answers

we will so
101 Points
6 years ago
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 of
rotation.
Therefore mass of that small area
m/a2 × a dx (m = mass of the square ; a = side of the plate)
I = a0
3
a
0
(m/ a2 ) ax2dx  (m/ a)(x / 3)]
= ma2/3
Therefore 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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