Problem 58
Question
A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of \(0.15 \mathrm{~m}\). (a) Through what distance does the tip move in one revolution? What are (b) the tip's speed and (c) the magnitude of its acceleration? (d) What is the period of the motion?
Step-by-Step Solution
Verified Answer
(a) \(0.30\pi \ m\), (b) \(6\pi \ m/s\), (c) \(240\pi^2 \ m/s^2\), (d) \(0.05 \ s\)
1Step 1: Determine the Distance per Revolution
The distance the tip moves in one revolution is the circumference of the circle formed by the blade. The formula for the circumference is \( C = 2\pi r \). Here \( r = 0.15 \ m \), so:\[C = 2 \pi \times 0.15 \ m = 0.30\pi \ m\]Thus, the distance the tip moves in one revolution is \(0.30\pi \ m\).
2Step 2: Calculate the Tip's Speed
The speed of the tip is the distance covered in one minute divided by the time. Since the fan completes 1200 revolutions per minute, the distance covered is:\[1200 \times 0.30 \pi \ m = 360\pi \ m\]The speed \( v \) is thus:\[v = \frac{360\pi \ m}{60 \ s} = 6\pi \ m/s\]
3Step 3: Calculate the Magnitude of Acceleration
The acceleration of the tip is centripetal, given by the formula \( a = \frac{v^2}{r} \). Using \( v = 6\pi \ m/s \) and \( r = 0.15 \ m \):\[a = \frac{(6\pi)^2}{0.15} = \frac{36\pi^2}{0.15} \ m/s^2\]Simplifying further:\[a = 240\pi^2 \ m/s^2\]
4Step 4: Determine the Period of the Motion
The period \( T \) is the time taken for one complete revolution. Since there are 1200 revolutions in one minute (60 seconds), the period is:\[T = \frac{60 \ s}{1200} = 0.05 \ s\]
Key Concepts
Centripetal AccelerationCircumference CalculationRevolution PeriodSpeed of Rotating Objects
Centripetal Acceleration
Centripetal acceleration is a key concept in rotational motion. It is the acceleration that keeps an object moving in a circular path. Instead of pushing the object outward, it keeps pulling the object towards the center of the circle. This means the direction of centripetal acceleration is always towards the center of the circle.
To calculate centripetal acceleration (\[a\]), we use the formula:
To calculate centripetal acceleration (\[a\]), we use the formula:
- \[a = \frac{v^2}{r}\],where:
- \(v\) is the speed of the object, and
- \(r\) is the radius of the circular path.
Circumference Calculation
Circumference is an essential measurement for circular paths. It gives us the total distance around a circle. For any rotating object, knowing the circumference helps us understand how far the object travels in one complete circle.
The formula for circumference (\[C\]) is:
The formula for circumference (\[C\]) is:
- \[C = 2\pi r\],where:
- \(r\) is the radius of the circle.
Revolution Period
The revolution period (\[T\]) is the time taken to complete one full cycle around a circular path. In rotational motion, the period helps us understand how quickly an object is revolving.
The formula to determine the period is:
The formula to determine the period is:
- \[T = \frac{1}{f}\],where:
- \(f\) is the frequency, the number of cycles per unit time.
Speed of Rotating Objects
The speed of a rotating object describes how fast the object moves along its circular path. It links the distance an object travels to the time taken. In rotational dynamics, it’s the speed at the circumference of the object's rotational path.
The formula for speed (\[v\]) is:
The formula for speed (\[v\]) is:
- \[v = \frac{d}{t}\],where:
- \(d\) is the distance traveled, and
- \(t\) is the time taken.
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