91Ó°ÊÓ

In problems involving rotational kinematics, angular speed (\(\omega\)) needs to often be converted between different standard units. Angular speed tells us how fast an object rotates or revolves relative to another point.
A common task is converting rotations per minute (rpm) to radians per second (rad/s). The conversion factor is derived from the fact that one complete revolution is equal to \(2\pi\) radians and a minute contains 60 seconds. Therefore, the conversion from rpm to rad/s is given by the formula:

• \(\omega = \text{rpm} \times \frac{2 \pi}{60}\)

For example, to convert 70 rpm to rad/s:

• First, multiply 70 by \(2\pi\), which gives approximately 439.82.
• Then, divide by 60 to get approximately 7.33 rad/s.

This converted angular speed value is crucial for further calculations in dynamics, such as determining torque and power.

The work-energy principle is a fundamental concept in physics that relates work done on a system to the change in energy of that system. In the context of rotational motion, it is often used to find the work done by a torque in driving a complete rotation.
In this scenario, the equation for work done (\(W\)) by a time-dependent torque \(\tau\) over time is:

• \(W = \int \tau \, dt\)

For a sinusoidal torque given by \(\tau = \tau_{0} \sin(\omega t)\), the definite integral from 0 to \(2\pi/\omega\) yields:

• \(W = \tau_{0} \cdot \frac{2}{\omega}\)

Using values from an example, maximum torque \(\tau_{0}\) equals 38.5 N·m, and angular speed \(\omega\) is 7.33 rad/s. Therefore, the work done in a complete pedal turn is:

• \(W = \frac{2 \cdot 38.5}{7.33}\approx 10.5 \text{ Joules}\)

This principle illustrates how forces culminating over time contribute energy to rotational movements.

Calculating power gives insight into how much energy is being transferred or transformed per unit of time. The average power in rotational motion is calculated using the work done and the period of the motion.
Average power (\(P\)) is defined by the equation:

• \(P = \frac{W}{T}\)

where \(W\) is the work completed in one cycle and \(T\) is the period of the cycle. To find the period \(T\), use:

• \(T = \frac{2\pi}{\omega}\)

In the given example:

• Work done \(W\) is 10.5 Joules, and \(\omega = 7.33 \text{ rad/s}\).
• The period \(T\) becomes \(\frac{2\pi}{7.33} \approx 0.86 \text{ seconds}\).
• Thus, \(P = \frac{10.5}{0.86} \approx 12.2 \text{ Watts}\).

This demonstrates how efficiently energy is converted into rotations, measuring the cyclist's exertion level in watts.