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Chapter 10: Problem 16

The driving wheel of a belt drive attached directly to an electric motor (as depicted in Fig. \(10-5\) ) has a diameter of \(38 \mathrm{~cm}\) and operates at \(1200 \mathrm{rpm}\). The motor turns the wheel, which moves the continuous looping belt, whose other end goes around a pulley, turning it and the shaft of some machine attached to it. The tension in the belt is \(130 \mathrm{~N}\) on the slack side and \(600 \mathrm{~N}\) on the tight side. Find the horsepower transmitted by the wheel to the belt and hence to the machine. Assume friction is negligible and there are no energy losses.

Short Answer

Expert verified
The horsepower transmitted is approximately 5.68 HP.

Step by step solution

01

Calculate the Torque

Torque (\( \tau \)) transmitted by the belt can be calculated using the formula \( \tau = (T_t - T_s) \times r \) where \( T_t \) is the tension on the tight side, \( T_s \) is the tension on the slack side, and \( r \) is the radius of the wheel. Given, \( T_t = 600 \, \text{N}, \, T_s = 130 \, \text{N}, \, \text{diameter} = 38 \, \text{cm} \Rightarrow r = 0.19 \, \text{m} \). Substituting values, we get \( \tau = (600 - 130) \times 0.19 \).
02

Convert RPM to Angular Velocity

The angular velocity \( \omega \) is given by \( \omega = \frac{2 \pi n}{60} \), where \( n \) is the rotational speed in rpm. Given \( n = 1200 \, \text{rpm} \), substituting we get \( \omega = \frac{2 \pi \times 1200}{60} \).
03

Calculate the Power in Watts

Power \( P \), in watts, transmitted by the wheel is given by \( P = \tau \times \omega \). Insert the values for \( \tau \) from Step 1 and \( \omega \) from Step 2 to find \( P \).
04

Convert Power to Horsepower

Convert the power calculated in Step 3 from watts to horsepower using the conversion \( 1 \, \text{HP} = 746 \, \text{W} \). Thus, horsepower \( \text{HP} = \frac{P}{746} \).

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Key Concepts

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

Torque Calculation
Torque is a fundamental concept in physics that quantifies the rotational force acting on an object. This measurement is especially crucial in mechanical systems that involve rotational motion, like gears and wheels.
To calculate torque, you use the difference between forces interacting with the rotational object and multiply it by the radius of the object. In our original exercise, we use:
  • The formula: \( \tau = (T_t - T_s) \times r \)
  • Where \( T_t \) is the force on the tight side, \( T_s \) is the force on the slack side, and \( r \) is the radius (half of the diameter).
Understanding torque is essential for engineering and physics applications because it helps in designing systems that can efficiently handle rotational forces.
Angular Velocity
The rotational speed of any object is captured by the term "angular velocity". It effectively tells us how quickly an object is moving in a circular path. When we talk about objects like wheels or pulleys, knowing their angular velocity helps us understand how effectively they are operating.
A vital part of solving our problem involves converting revolutions per minute (rpm) into angular velocity. This is done using the formula:
  • \( \omega = \frac{2 \pi n}{60} \)
  • Here, \( n \) represents the speed in rpm.
By understanding angular velocity, you gain insights into the speed and performance of machines, which is essential in optimizing mechanical systems.
Power Conversion
Converting power from one unit to another is frequent in physics, as it helps standardize calculations and communicate values universally. In the context of engines and motors, we often convert between watts and horsepower.
Power, in general, is the rate at which work is done or energy is transferred. The calculated torque and angular velocity from the previous concepts are multiplied as follows:
  • Power \( P \), in watts, is given by \( P = \tau \times \omega \).
  • To express this power in horsepower, recognizing the conversion \( 1 \text{ HP} = 746 \text{ W} \) is necessary.
  • Thus, \( \text{HP} = \frac{P}{746} \), offering a tangible measure of the machine's power.
Understanding power conversion is significant in both designing and comparing mechanical systems, ensuring that they are adequately powered for their required tasks.
Mechanical Systems
At the core of mechanical systems, like the belt drive outlined in the exercise, lies the interplay of forces, rotation, and power. Understanding each component and how they interreact is fundamental to engineering.
A mechanical system, in essence, involves:
  • Components such as pulleys, belts, motors, and shafts, which work together to perform a function.
  • The interaction of these parts to achieve desired outputs, which necessitates precise calculations and considerations.
  • The importance of identifying forces (like tension) and converting these into useable power through torque and angular velocity.
The ability to precisely analyze and improve these systems can enhance efficiency, reliability, and output, making an understanding of the basics crucial for aspiring engineers.

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Most popular questions from this chapter

An airplane propeller has a mass of \(70 \mathrm{~kg}\) and a radius of gyration of \(75 \mathrm{~cm}\). Find its moment of inertia. How large a torque is needed to give it an angular acceleration of \(4.0 \mathrm{rev} / \mathrm{s}^{2} ?\) $$ I=M k^{2}=(70 \mathrm{~kg})(0.75 \mathrm{~m})^{2}=39 \mathrm{~kg} \cdot \mathrm{m}^{2} $$ To be able to use \(\tau=I \alpha\), we must have \(\alpha\) in \(\mathrm{rad} / \mathrm{s}^{2}\) : $$ \alpha=\left(4.0 \frac{\mathrm{rev}}{\mathrm{s}^{2}}\right)\left(2 \pi \frac{\mathrm{rad}}{\mathrm{rev}}\right)=8.0 \pi \mathrm{rad} / \mathrm{s}^{2} $$ Then \(\tau=I \alpha=\left(39 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(8.0 \pi \mathrm{rad} / \mathrm{s}^{2}\right)=0.99 \mathrm{kN} \cdot \mathrm{m}\)

A \(25-\mathrm{kg}\) wheel has a radius of \(40 \mathrm{~cm}\) and turns freely on a horizontal axis. The radius of gyration of the wheel is \(30 \mathrm{~cm}\). A \(1.2\) -kg mass hangs at the end of a thin cord that is wound around the rim of the wheel. This mass falls and causes the wheel to rotate. Find the acceleration of the falling mass and the tension in the cord, whose mass can be ignored.

A disk like the lower one in Fig. \(10-11\) has a moment of inertia \(I=0.0150 \mathrm{~kg} \cdot \mathrm{m}^{2}\), and is turning at \(3.0 \mathrm{rev} / \mathrm{s}\). A trickle of sand falls onto the revolving disk at a distance of \(20 \mathrm{~cm}\) from the axis and builds a 20 -cm radius narrow ring of sand on it. How much sand must fall on the disk for it to slow to \(2.0 \mathrm{rev} / \mathrm{s} ?\) When a mass \(\Delta m\) of sand falls onto the disk, the moment of inertia of the disk is increased by an amount \(r^{2} \Delta m\), as shown in the preceding problem. After a mass \(m\) has fallen on the disk, the system's moment of inertia has increased to \(I+m r^{2}\). (Note how this agrees with the hoop in Fig.10-1.) Because the sand originally had no angular momentum, the law of conservation of momentum gives $$ \begin{array}{|ll} \text { (Momentum before) }=(\text { Momentum after }) & \text { or } \quad I \omega_{i}=\left(I+m r^{2}\right) \omega_{f} \end{array} $$ from which $$ m=\frac{I\left(\omega_{i}-\omega_{f}\right)}{r^{2} \omega_{f}}=\frac{\left(0.0150 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(6.0 \pi-4.0 \pi) \mathrm{rad} / \mathrm{s}}{\left(0.040 \mathrm{~m}^{2}\right)(4.0 \pi \mathrm{rad} / \mathrm{s})}=0.19 \mathrm{~kg} $$

Compute the rotational KE of a \(25-\mathrm{kg}\) wheel rotating at \(6.0 \mathrm{rev} / \mathrm{s}\) if the radius of gyration of the wheel is \(22 \mathrm{~cm}\).

The driving side of a belt has a tension of \(1600 \mathrm{~N}\), and the slack side has 500 - \(\mathrm{N}\) tension. The belt turns a pulley \(40 \mathrm{~cm}\) in radius at a rate of \(300 \mathrm{rpm} .\) This pulley drives a dynamo having 90 percent efficiency. How many kilowatts are being delivered by the dynamo?
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