Magnitude of Acceleration on Merry-Go-Round

  • Thread starter teresalin2004
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In summary, the question asks for the magnitude of acceleration of a 30 kg child on a 2.1 meter diameter merry-go-round that is pushed from rest to a final speed of 1.7 rotations per second in 25 seconds. Using radians, the angular acceleration is found to be 0.427 rad/s². However, the attempt at solving the question using rω² is incorrect as the diameter must be converted to radius and there is only one question to be answered.
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teresalin2004
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Homework Statement


At a playground you let your kid sister (30 kg) kneel on the surface (near the edge) of a 2.1 meter diameter merry-go-round. You push the merry-go-round so that it constantly accelerates from rest to a final speed of 1.7 rotations per second at the end of a 25 second interval. (please complete all questions using radians)
After it comes up to speed, you stop pushing and the merry-go-round continues to run at a constant rate of 1.7 rotations per second. What is the magnitude of your sister's acceleration, if any?

Homework Equations


V is the tangental velocity in m/s
r is radius of circle in meters
ω is angular velocity in radians/sec
a is in m/s²
[/B]
1.7 rotations per second x 2π rad/rev = 10.68 rad/s
ang. acc. = Δω/Δt = 10.68/25 = 0.427 rad/s²
134 rad x 1 rev/2πrad = 21.25 rev

The Attempt at a Solution


I got the first part of the question right, but I have no idea what I am doing wrong for the last part shown above. This is what I did:

a = rω² = (2.1)(10.68)² = 240 m/s²
[/B]
 
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  • #2
Diameter =/= radius!
 
  • #3
in the first place, 2.1 meters is the diameter, and you have to use the radius. In the second place, as far as I can see there is only one question.
 

Related to Magnitude of Acceleration on Merry-Go-Round

What is the definition of acceleration in relation to a merry-go-round?

Acceleration is the rate of change of velocity over time. In the context of a merry-go-round, it refers to the change in speed or direction of the riders as the ride rotates.

How does the speed of a merry-go-round affect the acceleration experienced by the riders?

The speed of a merry-go-round directly affects the acceleration experienced by the riders. The faster the ride spins, the greater the acceleration felt by the riders due to the change in velocity.

What factors contribute to the acceleration experienced on a merry-go-round?

The main factor contributing to acceleration on a merry-go-round is the rotation of the ride. Other factors such as the weight and position of the riders can also affect the acceleration experienced.

Is there a difference in acceleration between the outer and inner parts of a merry-go-round?

Yes, there is a difference in acceleration between the outer and inner parts of a merry-go-round. The outer part, which has a larger radius, has a greater linear speed and therefore experiences a greater acceleration compared to the inner part.

How can acceleration on a merry-go-round be calculated?

To calculate acceleration on a merry-go-round, the linear speed and angular velocity of the ride must be known. The formula for acceleration is a = rω^2, where a is acceleration, r is the radius of the ride, and ω is the angular velocity.

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