1.) A lawnmower engine claims to provide 18 hp of power and 6.25 ft lb of torque. If this is true, what must be the rotational speed in RPM?

2.) The conversion of the shaft power of a wind turbine to electricity is the function of the generator. Assume it has an efficiency of 0.45. If the wind turns the shaft of a generator at 26.2 rad/s (about 250 RPM) and the generator produces 300 W of electricity, what torque is produced by the wind on the shaft?

To answer these questions, we will need to use the formulas that relate power, torque, and rotational speed.

1.) To calculate the rotational speed in RPM, we can use the formula:

Rotational speed (RPM) = (Torque × 60) / (2π × Power)

Given that the power is 18 hp and the torque is 6.25 ft lb, we can convert them to SI units:

Power = 18 hp × 745.7 W/hp = 13,423.6 W
Torque = 6.25 ft lb × 1.356 Nm/ft lb = 8.496 Nm

Plugging these values into the formula, we get:

Rotational speed (RPM) = (8.496 Nm × 60) / (2π × 13,423.6 W)

Calculating this expression gives us the rotational speed in RPM.

2.) To calculate the torque produced by the wind on the shaft, we can use the formula:

Torque = Power / (Rotational speed × Efficiency)

Given that the power is 300 W, the rotational speed is 26.2 rad/s, and the efficiency is 0.45, we can plug these values into the formula:

Torque = 300 W / (26.2 rad/s × 0.45)

Calculating this expression gives us the torque produced by the wind on the shaft.