Problem 17 - Applied Physics

Question

A tangential force of \(150 \mathrm{~N}\) is applied to a flywheel of diameter \(45 \mathrm{~cm}\) to maintain a constant angular velocity of 175 rpm. How much work is done per minute?

Step-by-Step Solution

Answer

37117 Joules per minute.

Step 1: Convert Diameter to Radius

To find the work done, we first need the radius of the flywheel. The diameter is given as \(45\,\text{cm}\). The radius \(r\) is half of the diameter, so: \[ r = \frac{45}{2} = 22.5\,\text{cm} = 0.225\,\text{m} \] (convert to meters for consistency in SI units).

Step 2: Calculate the Torque

Next, we calculate the torque \(\tau\) using the formula \(\tau = r \times F\), where \(F\) is the tangential force. Given \(F = 150\,\text{N}\) and \(r = 0.225\,\text{m}\), the torque is:\[ \tau = 0.225\,\text{m} \times 150\,\text{N} = 33.75\,\text{Nm} \]

Step 3: Convert Angular Velocity to Radians per Second

The angular velocity is initially given in revolutions per minute (rpm). To convert it to radians per second, we use the conversion: \[ 175\,\text{rpm} \times \frac{2\pi\,\text{radians}}{1\,\text{revolution}} \times \frac{1\,\text{minute}}{60\,\text{seconds}} \] This gives:\[ \omega = \frac{175 \times 2 \pi}{60} \approx 18.33\,\text{radians per second} \]

Step 4: Calculate the Work Done per Minute

Work done due to rotation can be calculated using the formula:\[ W = \tau \cdot \theta \] where \(\theta\) is the angular displacement in radians. Since we want the work done in one minute, the angular displacement \(\theta\) is:\[ \theta = \omega \times \text{time} = 18.33\,\text{radians/second} \times 60\,\text{seconds} = 1099.8\,\text{radians} \]Thus, the work done \(W\) is:\[ W = 33.75\,\text{Nm} \times 1099.8\,\text{radians} = 37117.25\,\text{Joules} \] which rounds to approximately \(37117\,\text{Joules} \).

Key Concepts

Torque Calculation

Torque is a measure of the rotational force applied around an axis. To calculate torque, use the formula:

• \( \tau = r \times F \)

Here, \( \tau \) is the torque in Newton-meters (Nm), \( r \) is the radius in meters, and \( F \) is the tangential force in Newtons. In the given problem, applying a force of 150 N tangentially to a flywheel with a 22.5 cm radius leads to a torque:

• First, convert the radius to meters: \( 22.5\,\text{cm} = 0.225\,\text{m} \)
• Next, calculate the torque: \( \tau = 0.225\,\text{m} \times 150\,\text{N} = 33.75\,\text{Nm} \)

This value represents the effort needed to keep the flywheel in constant motion, overcoming any resisting forces. Understanding how torque functions are essential in fields like automotive engineering, where it plays a critical role in engine performance and efficiency.

Angular Velocity Conversion

Angular velocity reflects how fast something rotates and is typically measured in radians per second (rad/s). Converting from revolutions per minute (rpm) to rad/s involves the following steps:
Multiply the rpm by the conversion factors:

• Use the fact that one revolution equals \( 2\pi \) radians.
• Convert minutes to seconds by dividing by 60.

The calculation looks like this:

• \( 175\,\text{rpm} \times \frac{2 \pi\,\text{radians}}{1\,\text{revolution}} \times \frac{1\,\text{minute}}{60\,\text{seconds}} \)
• \( \omega = \frac{175 \times 2 \pi}{60} \approx 18.33\,\text{radians per second} \)

This conversion is crucial for handling equations in rotational motion, as radians offer a direct proportionality to arc length and are SI unit compliant.

Work Done in Rotational Motion

In rotational motion, work done is calculated using torque and angular displacement. The formula used is:

• \( W = \tau \cdot \theta \)

Here, \( W \) is the work in Joules, \( \tau \) is the torque in Nm, and \( \theta \) refers to the angular displacement in radians. For the given problem:

• First, find angular displacement: \( \theta = \omega \times \text{time} = 18.33 \times 60 = 1099.8\,\text{radians} \)
• Next, calculate work done: \( W = 33.75\,\text{Nm} \times 1099.8\,\text{radians} = 37117.25\,\text{Joules} \)

This result tells us the energy required to maintain the flywheel's motion every minute. Understanding the concept of work in rotational systems is indispensable for designing machines that incorporate spinning parts, ensuring they operate efficiently without unnecessary energy wastage.