How Many Revolutions Does a 25 rad/s Wheel Turn in 10 Seconds?

Understanding rotational motion is crucial in various fields, including physics, engineering, and robotics. This article will guide you through a practical example to determine how many revolutions a wheel with a rotational speed of 25 radians per second (rad/s) makes in 10 seconds. We will break down the problem into two key steps: calculating the total radians turned in 10 seconds and then converting those radians into revolutions.

Step 1: Calculating the Total Radians Turned in 10 Seconds

Given that the wheel rotates at 25 radians every second, we need to find out how many radians it will rotate in 10 seconds. This can be calculated by simply multiplying the rotational speed (25 rad/s) by the time (10 seconds).

The total radians turned in 10 seconds is:

[ 25 text{ rad/s} times 10 text{ s} 250 text{ radians} ]

Step 2: Converting Radians into Revolutions

To understand how many revolutions the wheel completes, we need to convert the total radians into revolutions. It is important to know that one complete revolution is equivalent to 2π radians. Therefore, we need to divide the total radians (250 radians) by 2π to find out the number of revolutions.

The number of revolutions (N) can be calculated using the formula:

[ N frac{250 text{ radians}}{2π text{ radians/revolution}} ]

Let's simplify this expression:

[ N frac{250}{2π} approx 39.79 ]

Therefore, the wheel will turn approximately 39.79 revolutions in 10 seconds.

Conclusion

By following these two simple steps, we can accurately determine the number of revolutions a wheel with a rotational speed of 25 rad/s will make in 10 seconds. This problem highlights the importance of understanding units and conversions in rotational motion, which are fundamental concepts in physics and engineering.

Related Keywords

revolution radians rotational motion

References

1. Feynman, R. P. (1963). The Feynman Lectures on Physics: Vol. 1, Mainly Mechanics, Radiation, and Heat. Basic Books.

2. Morse, P. M., Feshbach, H. (1953). . McGraw-Hill.