Question

Arrangement shown below is used to measure the power of a rotating shaft having 25 revolutions per second. If the spring opposing force as measured on the scale was found to be 105 N then the power consumed by the shaft in kW is ____.

• A. 4.71

Answer 1

The correct option is A 4.71
$\text{The net force acting on the shaft}$

$F_{net}=F_s-F_w$

$F_s=\text{Spring force},F_w=\text{Force due to gravity}$

$\therefore F_{net}=105-m\times g$

$=105-0.5\times 9.8$

$=100.1N$

$\therefore \text{Torque}=F_{net}\times R$

$R=\frac{D}{2}=\frac{2ft}{2}=1ft={12}^{\prime \prime }=30.0cm$

$\therefore \text{Torque}=100.1\times 0.30=30.03N-m$

$\therefore \text{The power consumed by the shaft}$

$P=2\pi NT$

$=2\pi \times 25\times 30.03$

$=4717.10Watt$

$P=4.71kW$