/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Crankshaft An automobile cranksh... [FREE SOLUTION] | 91Ó°ÊÓ

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Crankshaft An automobile crankshaft transfers energy from the engine to the axle at the rate of \(100 \mathrm{hp}(=74.6 \mathrm{~kW})\) when rotating at a speed of 1800 rev/min. What torque (in newtonmeters) does the crankshaft deliver?

Short Answer

Expert verified
The torque delivered is 396 Nm.

Step by step solution

01

Convert Power from Horsepower to Watts

First, convert the given power from horsepower (hp) to watts (W). Given that 1 hp = 746 W, we have the power as:\[ P = 100 \text{ hp} \times 746 \text{ W/hp} = 74600 \text{ W} \]
02

Convert Rotational Speed to Radians per Second

Next, convert the given rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). Use the conversion factor: \[ 1 \text{ rev} = 2\text{Ï€ rad} \text{ and } 1 \text{ min} = 60 \text{ s} \]Calculate:\[ \text{Rotational Speed} = 1800 \text{ rev/min} \times \frac{2Ï€ \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 188.4 \text{ rad/s} \]
03

Use the Power-Torque Relationship

The relationship between power (P), torque (τ), and angular velocity (ω) is given by the formula: \[ P = τ \times ω \]Rearrange this formula to solve for torque (τ): \[ τ = \frac{P}{ω} \]
04

Calculate the Torque

Substitute the known values for power (P) and angular velocity (ω) into the torque equation:\[ τ = \frac{74600 \text{ W}}{188.4 \text{ rad/s}} = 396 \text{ Nm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Conversion
The exercise begins with converting the power from horsepower (hp) to watts (W). This is necessary because standard scientific formulas typically require power to be in watts. Power in horsepower is commonly used in automotive and mechanical fields, but we need it in watts to compute torque correctly. Given that 1 hp equals 746 watts, we multiply the power in hp by 746:
\[ P = 100 \text{ hp} \times 746 \text{ W/hp} = 74600 \text{ W} \]
This conversion helps us to align the power units with the SI system, making further calculations more straightforward.
Rotational Speed
The next step involves converting the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). This is crucial because the relationship between power and torque requires angular speed in radians per second. The conversion process includes two key conversions:
  • 1 revolution = 2Ï€ radians
  • 1 minute = 60 seconds
Using these conversions, we calculate rotational speed as:
\[ \text{Rotational Speed} = 1800 \text{ rev/min} \times \frac{2Ï€ \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 188.4 \text{ rad/s} \]
This calculation is important because using radians per second keeps our formulas consistent with angular measurements in the SI unit system.
Torque Formula
The most critical part of this exercise is applying the torque formula, which involves the relationship between power (P), torque (τ), and angular velocity (ω). This relationship is captured by the formula:
\[ P = τ \times ω \]
To isolate and solve for torque (Ï„), we rearrange the formula as:
\[ τ = \frac{P}{ω} \]
We can substitute the known values of power (74600 W) and angular velocity (188.4 rad/s) into the equation:
\[ Ï„ = \frac{74600 \text{ W}}{188.4 \text{ rad/s}} = 396 \text{ Nm} \]
The torque calculated here indicates the twisting force the crankshaft exerts, measured in newton-meters (Nm). Understanding this relationship helps to analyze and design components in mechanical systems, ensuring they function efficiently under specific power and speed conditions.

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Most popular questions from this chapter

Torque on a Certain Wheel A torque of \(32.0 \mathrm{~N} \cdot \mathrm{m}\) on a certain wheel causes a rotational acceleration of \(25.0 \mathrm{rad} / \mathrm{s}^{2} .\) What is the wheel's rotational inertia?

Small Ball A small ball of mass \(0.75 \mathrm{~kg}\) is attached to one end of a \(1.25\) -m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is \(30^{\circ}\) from the vertical, what is the magnitude of the torque about the pivot?

Meter Stick Held Vertically A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.)

Bicycle Pedal Arm The length of a bicycle pedal arm is \(0.152 \mathrm{~m}\), and a downward force of \(111 \mathrm{~N}\) is applied to the pedal by the rider's foot. What is the magnitude of the torque about the pedal arm's pivot point when the arm makes an angle of (a) \(30^{\circ}\), (b) \(90^{\circ}\). and (c) \(180^{\circ}\) with the vertical?

Flywheel Rotating A flywheel with a diameter of \(1.20 \mathrm{~m}\) has a rotational speed of 200 rev/min. (a) What is the rotational speed of the flywheel in radians per second? (b) What is the translational speed of a point on the rim of the flywheel? (c) What constant rotational acceleration (in revolutions per minute-squared) will increase the wheel's rotational speed to 1000 rev/min in 60 s? (d) How many revolutions does the wheel make during that \(60 \mathrm{~s}\) ?

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