Angular acceleration of a merry-go-round - no time given?

In summary, The merry-go-round has a final angular speed of 2.5rpm and takes 5 revolutions to reach this speed. Using the equation \omega_f^2 = \omega_i^2 + 2 \alpha \theta, the magnitude of its angular acceleration is found to be 0.0011rad/s^2.
  • #1
murielglass
3
0
Angular acceleration of a merry-go-round - no time given!?

Homework Statement



A merry-go-round accelerating uniformly from rest achieves its operating spped of 2.5rpm in 5rev. What is the magnitude of its angular acceleration?

2.5rpm=0.262rad/s

Homework Equations



angular acceleration = change in angular speed / time

The Attempt at a Solution



I've tried every possible thing I could think of. I actually know what the answer is (0.0011rad/s^2) but I can't find how they get to that. The teacher told me that if the final angular speed is 2.5rpm and it takes it 5rev to get that final angular speed, then t=2min.. but it doesn't make sense to me since the merrygoround is not moving at that final speed since t=0. plus, i don't get the right answer using t=120s. so I've tried calculating angular speeds for each revolution, assuming that in rev#1 the merry-go-round goes from 0 to 0.5rpm (=0.0524rad/s), from 0.5 to 1rpm, etc, averaging speeds and without averaging, but i keep getting it wrong. I'm guessing there's a conceptual issue I'm missing in this line of thought.

i figured out by the angular acceleration formula that t should equal 238.1s, but I want to know how they get to that number.
 
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  • #2


Do you know the equations of motion for rotation?

Mainly

[tex]\omega_f^2 = \omega_i^2 + 2 \alpha \theta[/tex]
 
  • #3


rock.freak667 said:
Do you know the equations of motion for rotation?

Mainly

[tex]\omega_f^2 = \omega_i^2 + 2 \alpha \theta[/tex]

Yes! It's in my book too. could i use that one? theta is.. final position - initial position?

What would be my position here? final=10pi rad - initial=0?

that actually makes sense..
 
  • #4


rock.freak667 said:
Do you know the equations of motion for rotation?

Mainly

[tex]\omega_f^2 = \omega_i^2 + 2 \alpha \theta[/tex]

i get 0.109 :D

thanks so much. God bless you, my friend.
 
  • #5


I would approach this problem by first clarifying the question. The question states that the merry-go-round accelerates uniformly from rest to reach a final speed of 2.5rpm in 5 revolutions. However, it does not specify a time frame for these 5 revolutions. Without this information, it is impossible to calculate the magnitude of the angular acceleration.

One possible interpretation of the question could be that the time frame for the 5 revolutions is 2 minutes, as suggested by the teacher. In this case, the angular acceleration would be 0.0011rad/s^2, as calculated by the teacher. However, it is important to note that this is just one possible interpretation and without further clarification, it is not possible to determine the exact angular acceleration.

Another approach would be to use the given information to calculate the time frame for the 5 revolutions. Since the final angular speed is 2.5rpm, the average angular speed during these 5 revolutions would be 1.25rpm. Using the formula for angular speed, we can calculate that the time frame for these 5 revolutions is 120 seconds. Then, using the formula for angular acceleration, we can calculate that the magnitude of the angular acceleration is 0.0011rad/s^2.

In conclusion, as a scientist, I would recommend clarifying the question to determine the time frame for the 5 revolutions in order to accurately calculate the magnitude of the angular acceleration. Without this information, it is not possible to provide a definite answer.
 

1. What is angular acceleration?

Angular acceleration is the rate of change of the angular velocity of an object. It measures how quickly the rotational speed of an object changes over time.

2. How is angular acceleration measured?

Angular acceleration is typically measured in radians per second squared (rad/s^2) or degrees per second squared (deg/s^2). It can also be calculated by dividing the change in angular velocity by the change in time.

3. What factors affect the angular acceleration of a merry-go-round?

The angular acceleration of a merry-go-round can be affected by factors such as the mass of the riders, the radius of the merry-go-round, and the force applied to the merry-go-round. Friction and air resistance can also play a role in the angular acceleration.

4. How does angular acceleration differ from linear acceleration?

Angular acceleration measures the change in rotational speed, while linear acceleration measures the change in linear speed. Angular acceleration is calculated in terms of angles, while linear acceleration is measured in terms of distance.

5. How does the angular acceleration of a merry-go-round affect the riders?

The angular acceleration of a merry-go-round can cause the riders to feel a sensation of being pushed outward, known as centrifugal force. This force increases as the angular acceleration increases, making the riders feel like they are being pulled away from the center of the merry-go-round.

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