Question

An automobile crankshaft transfers energy from the engine to the axle at the rate of $\mathbf{100}\mathbf{\text{\hspace{0.17em}hp}}\mathbf{}\mathbf{(}\mathbf{=}\mathbf{}\mathbf{74}\mathbf{.6}\mathbf{}\mathbf{\text{\hspace{0.17em}kW}}\mathbf{)}$ when rotating at a speed of $\mathbf{1800}\mathbf{\text{\hspace{0.17em}rev}}\mathbf{/}\mathbf{\text{min}}$. What torque (in newton-meters) does the crankshaft deliver?

Short answer

Torque delivered by the crankshaft is$396\text{\hspace{0.17em}N.m}$ .

Step-by-step solution

Understanding the given information

1. Power of an automobile crankshaft is,$\mathrm{P}=100\text{\hspace{0.17em}hp}=74.6\text{\hspace{0.17em}kW}=74600\text{\hspace{0.17em}W}$.
2. The speed is, $\mathrm{f}=1800\text{\hspace{0.17em}rev}/\text{min}$.

Concept and formula used in the given question

You can find thetorque delivered by the crankshaft using the relation between torque, power, and angular speed.

$\mathbf{P}\mathbf{=}\mathbf{\tau \omega }$

Calculation for the torque the crankshaft delivers 

Frequency of the rotation of an automobile is

$\begin{array}{c}\text{f}=1800\frac{\text{rev}}{\text{min}}\times \frac{1\text{\hspace{0.17em}min}}{60\text{\hspace{0.17em}s}}\\ =30\text{\hspace{0.17em}rev}/\text{s}\end{array}$

Angular speed ofan automobile is

$\begin{array}{c}\mathrm{\omega }=2\mathrm{\pi f}\\ =2\left(3.142\right)(30\text{\hspace{0.17em}rev}/\text{s})\\ =188.52\text{\hspace{0.17em}rad}/\text{s}\end{array}$

Power of an automobile crankshaft is

$\begin{array}{c}\mathrm{P}=\mathrm{\tau \omega }\\ \mathrm{\tau }=\frac{\mathrm{P}}{\mathrm{\omega }}\\ =\frac{74600\text{\hspace{0.17em}W}}{188.52\text{\hspace{0.17em}rad}/\text{s}}\\ =395.71\text{\hspace{0.17em}N.m}\\ \simeq 396\text{\hspace{0.17em}N.m}\end{array}$

Therefore, the torque delivered by the crankshaft is $396\text{\hspace{0.17em}N.m}$.