On the application of a constant torque, a wheel is turned from rest to 400 rad in 10 s . If same torque continues to act, the angular velocity of the wheel after 20 s from the start is

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Aicharya · Accepted Answer

By using the following equation, &#952; = &#969; 0 t + 1 2 &#945; t 2 &#945; = 8 &#160; rad &#160; s - 2 Also ,&#160; &#969; = &#969; 0 + &#945;t Here &#969; 0 = 0 , &#945; = 8 &#160; rad &#160; s - 2 and t = 20 &#160; s &#8756; &#969; = 0 + 8 &#215; 20 = 160 rad s - 1

On the application of a constant torque, a wheel is turned from rest to 400 rad in 10 s . If same torque…

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