Circular rotational motion (kinematics)

In summary, the wheel starts from rest and accelerates uniformly to 200rpm in 6 seconds. After rotating at this speed for some time, the brakes are applied for 5 seconds and it takes a total of 3100 rotations. By calculating the rotations in the first and last intervals, we can find the rotations in the middle interval and use this to find the total time of rotation to be approximately 16.62 minutes. However, there are discrepancies in the problem and the book, leading to a possible typo in the book.
  • #1
dunn
18
0

Homework Statement



A wheel starts from rest and accelerates uniformly to 200rpm in 6s. After it has been rotating for some time at this speed, the brakes are applied and it takes 5 minutes to stop the wheel. If the total number of revolutions is 3100, calculate the total time of rotation.

Homework Equations



α = dω/dt
ω = dθ/dt

The Attempt at a Solution



Okay, so I need to calculate the number of rotations it completes in the first interval (between 0 and 6 seconds). I have the final angular velocity (ω) at the end of the first interval, so I need to find the angular acceleration (α) in order to calculate the number of radians (θ).

Since α is uniform, I integrate α = dω/dt and derive the following:

α(t-t0) = ω - ω0 (1)

Since t0 and ω0 = 0 at the start, I have:

α(t) = ω

At t = 6 and with ω = 8π/3 rad/s (200rpm), my angular acceleration α = 4π/9 rad/s2

Now I have ω = dθ/dt so α(t-t0) = ω - ω0 becomes:

α(t-t0) + ω0 = dθ/dt

Integrating this yields:

1/2α(t-t0)2 + ω0(t-t0) = θ - θ0 (2)

Putting in the relevant values (t = 6, α = 4π/9, ω0 = 0) I obtain θ = 8π rad or 4 rotations.

Now I need to calculate the last interval when the brakes are applied. Using equation (1) I obtain an angular acceleration α = -8π/900 rad/s2.

Now I put the relevant information into equation 2 (t = 300, ω0 = 8π/3 rad/s, α = -8π/900 rad/s2 I get θ = 400π or 200 rotations.

Since the wheel rotates 3100 times in total, I subtract the rotations from the first and last interval to obtain 3100 - 200 - 4 = 2198 rotations that occur during the middle interval. Since 2198 rotations = 5792 radians and ω is constant, I can integrate ω = dθ/dt to obtain:

ω(t-t0) = θ - θ0

or, more relevantly:

ωt = θ

Since I know ω and θ I divide 5792 rad by 8π/3 rad/s and get 691.4 seconds. Adding the times from the other intervals (6s and 300s) I get a total time of 997.4 seconds or ~16.62 minutes.

The problem is that the book says the answer should be 15.6 minutes. That would mean that the wheel would complete 3120 rotations in 15.6 minutes if it started at its full speed of 200rpm and never slowed down. That simply doesn't seem right.

I want to be sure that this is just an error in the book, so I'm posting this here in the hope that someone can look over my work and make sure I did everything correctly. Am I doing something wrong or is this just a typo?
 
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  • #2
200 rpm = 200*2π/60 rad/s = ...?
 
  • #3
rl.bhat said:
200 rpm = 200*2π/60 rad/s = ...?

:redface: I hate calculators that can't do fractions...

Thank you!
 
  • #4
Well, after playing with the problem more I came to the conclusion that there is in fact a typo in the book. The brakes are supposed to be applied for a total of 5 seconds, not 5 minutes. My initial math was wrong but this was further complicated by the fact that my correct math was rendering the wrong answer because the book has (yet another!) typo. I need a new book. :cry:
 
  • #5




Based on your calculations, it appears that you have approached the problem correctly. However, it is possible that there is a typo in the book or that the problem is poorly worded, leading to confusion and incorrect answers. As a scientist, it is important to critically examine all sources of information and question any discrepancies or inconsistencies. In this case, it may be helpful to double check your calculations and assumptions, and also consider reaching out to the author or instructor for clarification. Overall, your approach to the problem is sound and it is likely that the discrepancy lies in the given information rather than your calculations.
 

1. What is circular rotational motion?

Circular rotational motion, also known as circular motion, is a type of motion where an object moves along a circular path around a fixed point or axis. It involves both translational and rotational motion.

2. What is the difference between circular and linear motion?

The main difference between circular and linear motion is the path of the object. In circular motion, the object moves along a circular path, while in linear motion, the object moves along a straight path. Additionally, circular motion involves both translational and rotational motion, whereas linear motion only involves translational motion.

3. How is centripetal force related to circular rotational motion?

Centripetal force is the force that keeps an object moving in a circular path. In circular rotational motion, the centripetal force is responsible for constantly changing the direction of the object's velocity, keeping it in a circular path.

4. What is angular velocity in circular rotational motion?

Angular velocity is a measure of how fast an object is rotating around an axis. It is the rate of change of angular displacement with respect to time. It is usually measured in radians per second.

5. How is circular rotational motion related to the concept of inertia?

Inertia is the tendency of an object to resist changes in its motion. In circular rotational motion, the concept of inertia is seen in the object's resistance to changes in its rotational motion, which is described by its moment of inertia. The greater the moment of inertia, the greater the object's resistance to changes in its rotational motion.

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