Problem 24
Question
A merry-go-round makes 24 revolutions in a 3.0 -min ride. (a) What is its average angular speed in rad/s? (b) What are the tangential speeds of two people \(4.0 \mathrm{~m}\) and \(5.0 \mathrm{~m}\) from the center, or axis of rotation?
Step-by-Step Solution
Verified Answer
(a) 0.8378 rad/s; (b) 3.35 m/s at 4.0 m, 4.19 m/s at 5.0 m.
1Step 1: Convert time to seconds
First, convert the time from minutes to seconds. Since there are 60 seconds in a minute, the total time for the ride is \(3.0 \times 60 = 180\) seconds.
2Step 2: Calculate average angular speed
To find the average angular speed, we use the formula \(\omega = \frac{\theta}{t}\), where \(\omega\) is the angular speed, \(\theta\) is the angle in radians (total revolutions), and \(t\) is the time in seconds. Given 24 revolutions, we convert it to radians: \(24 \times 2\pi = 48\pi\) radians. Therefore, \(\omega = \frac{48\pi}{180}\, \text{rad/s} \approx 0.8378\, \text{rad/s}\).
3Step 3: Calculate tangential speed at 4.0 m
Tangential speed \(v\) is calculated using \(v = r \omega\), where \(r\) is the radius. At 4.0 m, \(v = 4.0 \times 0.8378 \approx 3.3512\, \text{m/s}\).
4Step 4: Calculate tangential speed at 5.0 m
Similarly, for a radius of 5.0 m, \(v = 5.0 \times 0.8378 \approx 4.1890\, \text{m/s}\).
Key Concepts
Tangential SpeedRevolutions to Radians ConversionPhysics Problem Solving StepsRotational Motion
Tangential Speed
Tangential speed is the linear speed of a point located on a rotating object. It is crucial in understanding motion in a circular path. For a point at a distance from the center of rotation, tangential speed is directly proportional to both the distance (radius) and the angular speed.
If we consider a merry-go-round, the tangential speed of a person riding it affects how quickly they move around the circular path.
If we consider a merry-go-round, the tangential speed of a person riding it affects how quickly they move around the circular path.
- The relationship is expressed by the formula: \( v = r \omega \), where \( v \) is the tangential speed, \( r \) is the radius distance from the center, and \( \omega \) is the angular speed.
- If you double the distance from the center, the tangential speed also doubles, illustrating why people sitting further out on a merry-go-round move faster than those close to the center.
Revolutions to Radians Conversion
The conversion from revolutions to radians is an essential step in solving rotational motion problems. A full circle in rotational motion signifies one complete revolution around the circular path, equal to \(2\pi\) radians.
The conversion process is crucial for computations involving angular speed, as most formulas in physics use radians rather than revolutions.
The conversion process is crucial for computations involving angular speed, as most formulas in physics use radians rather than revolutions.
- To convert revolutions to radians, multiply the number of revolutions by \(2\pi\). For example, 24 revolutions can be converted as follows: \(24 \times 2\pi = 48\pi \) radians.
- This allows calculations to proceed smoothly, using radians, which is standard in scientific computations.
Physics Problem Solving Steps
Solving physics problems effectively often involves a structured problem-solving approach. Let's break it down using our merry-go-round example:1. **Understand the Problem**: Recognize what quantities are given and what needs to be found. Here, you are given revolutions and time, and asked to find angular and tangential speeds.
2. **Convert Units**: Convert any non-standard units to suitable ones. For instance, convert time from minutes to seconds for synchronicity with calculations in the standard Physics unit.
3. **Use Formulas Effectively**: Apply relevant physics formulas. For angular speed, use \( \omega = \frac{\theta}{t} \). For tangential speed, try \( v = r \omega \).
By consistently following these steps, solving even complex physics problems becomes significantly more manageable.
2. **Convert Units**: Convert any non-standard units to suitable ones. For instance, convert time from minutes to seconds for synchronicity with calculations in the standard Physics unit.
3. **Use Formulas Effectively**: Apply relevant physics formulas. For angular speed, use \( \omega = \frac{\theta}{t} \). For tangential speed, try \( v = r \omega \).
- Ensure you’ve converted all necessary units, such as revolutions to radians, before applying these formulas.
By consistently following these steps, solving even complex physics problems becomes significantly more manageable.
Rotational Motion
Rotational motion refers to the motion of an object around a center or axis. This concept includes essential elements like angular speed, tangential speed, and total rotations.Angular speed refers to how fast an object rotates. It's the angle covered per unit of time, expressed in radians per second. In our merry-go-round exercise, the angular speed is determined by how many radians it covers in a minute.
Tangential speed describes how fast a point on the object moves along its circular path. It's different for points at different distances from the axis of rotation due to varying radii.
Tangential speed describes how fast a point on the object moves along its circular path. It's different for points at different distances from the axis of rotation due to varying radii.
- Both tangential and angular speeds are related, with tangential speed being calculated by multiplying angular speed by the radius (\(v = r\omega\)).
- This relationship explains why the same angular speed results in different tangential speeds at different radial positions.
Other exercises in this chapter
Problem 22
Determine which has the greater angular speed: particle \(A\), which travels \(160^{\circ}\) in \(2.00 \mathrm{~s}\), or particle \(\mathrm{B}\), which travels
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The tangential speed of a particle on a rotating wheel is \(3.0 \mathrm{~m} / \mathrm{s}\). If the particle is \(0.20 \mathrm{~m}\) from the axis of rotation, h
View solution Problem 26
The Earth rotates on its axis once a day and revolves around the Sun once a year. (a) Which is greater, the rotating angular speed or the revolving angular spee
View solution Problem 27
A little boy jumps onto a small merry-go-round (radius of \(2.00 \mathrm{~m}\) ) in a park and rotates for 2.30 s through an arc length distance of \(2.55 \math
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