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A shaft rotating at 1800 rpm is transmitting a power of 6.28 kW. The magnitude of the driving torque is

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42. 8 Nm

33.3 Nm

36.2 Nm

48.2 Nm

answer is B.

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Detailed Solution

$ω = 1800 rpm = \frac{2 π \times 1800}{60} = 60 π rad / s$

$P = τ ω \Rightarrow 6.28 \times 10^{3} = τ \times 60 \times 3.14$

$\Rightarrow τ = \frac{6.28 \times 10^{3}}{60 \times 3.14} = \frac{1000}{30} = 33.3 Nm$

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