Question

Calculate the torque developed by an airplane engine whose output is 2000 HP at an angular velocity of 2400 rev/min.

Answer 1

Hint:We know that work done by a constant torque being applied on a body is equal to the product of torque and angular displacement. If we can find the work done, we can find the power developed by dividing the work with time. So we have been given power in terms of horsepower and we have been given angular velocity which we can use to find angular displacement. Let’s take this slowly, one step at a time.

Complete step-by-step solution:
Firstly, converting power into the SI unit,
$ \Rightarrow P = 2000HP = 2000 \times 746J$SS [ $1HP = 746J$ ]
Now let’s convert our angular velocity from rev/min to rad/sec
\[\omega = 2400rev/\min = \dfrac{{2400 \times 2\pi }}{{60}}rad/s = 80\pi rad/s\]
Work done, \[W = \tau \times \Delta \theta \]
Differentiating both sides with respect to time, we get \[\dfrac{{dW}}{{dt}} = \tau \times \dfrac{{d\theta }}{{dt}} \Rightarrow P = \tau \times \omega \]
Substituting all the values in the above-obtained equation and calculating torque,
\[\tau = \dfrac{{2000 \times 746J}}{{80\pi }} = 5937N.m\]

Additional Information: Both torque and work have the same dimensional formula. The more torque an engine produces, the greater is its ability to perform work. The difference in both is that torque is a vector that produces angular acceleration about an axis whereas work is a scalar which indicates the increase in energy of an object under acceleration along a path.

Note:- Alternatively, we can approach this question as follows:
We know that power is equal to the product of force and velocity. Velocity is nothing but the product of radius and angular velocity, which will again lead us to the same conclusion that power is equal to torque multiplied by angular velocity. \[P = F \times v = (F \times r) \times \omega = \tau \times \omega \]