Calculating Maximum Speed and Acceleration for an Amusement Park Ride

[veronicak5678]

Homework Statement

An amusement park ride carries riders in a horizontal circle with radius 5m.

1) If the centripetal acc. is limited to .4 g for safety, what is the max tangential speed?
2) If the tangential speed is doubled, what is the new acceleration?

Homework Equations

A= v^2/r

The Attempt at a Solution

1) (.4g(5m))^(1/2)
max speed = 1.41 m/s

2) A = 2root2 m/s / 5m = 1.6 m/s^2

Does this look right?

 

[LowlyPion]

Homework Statement

An amusement park ride carries riders in a horizontal circle with radius 5m.

1) If the centripetal acc. is limited to .4 g for safety, what is the max tangential speed?
2) If the tangential speed is doubled, what is the new acceleration?

Homework Equations

A= v^2/r

The Attempt at a Solution

1) (.4g(5m))^(1/2)
max speed = 1.41 m/s

2) A = 2root2 m/s / 5m = 1.6 m/s^2

Does this look right?

Yep.

 

[veronicak5678]

OK!
I wasn't sure because the next problem refers to a bike traveling around a circular curve and asks for acceleration. I was going to do the same thing, but there's a note next to the question that says "Remember this is a vector!". Why is that a vector and not this?

 

[alphysicist]

Homework Statement

An amusement park ride carries riders in a horizontal circle with radius 5m.

1) If the centripetal acc. is limited to .4 g for safety, what is the max tangential speed?
2) If the tangential speed is doubled, what is the new acceleration?

Homework Equations

A= v^2/r

The Attempt at a Solution

1) (.4g(5m))^(1/2)
max speed = 1.41 m/s

I don't believe this is correct; I think you forgot to multiply by the factor of g.

2) A = 2root2 m/s / 5m = 1.6 m/s^2

Does this look right?

The acceleration started out as (0.4 g) which is about 4m/s^2, and increasing the speed will not make the acceleration decrease.

 

[veronicak5678]

I see. Using 9.8 m/s^2 for g, I get 19.6^(1/2) m/s for part 1 and 15.68 m/s^2 for part 2.
Still don't understand the next question. How can I calculate acceleration on a curve as a vector?

 

[LowlyPion]

I see. Using 9.8 m/s^2 for g, I get 19.6^(1/2) m/s for part 1 and 15.68 m/s^2 for part 2.
Still don't understand the next question. How can I calculate acceleration on a curve as a vector?

He's right. I missed the g. Sorry.

As to the acceleration that's given by v²/r but it's radially directed. If the tangential speed is also accelerating then the value of the tangential acceleration is a vector that is added to the radially directed centripetal acceleration.The resultant vector is then directed at an angle to the radius.

Btw: the first one is a vector too. It's radial. The question though was only concerned with its magnitude.

 

[veronicak5678]

Um, not sure I get all that.
This problem deals with constant speed, so I assume the tangential speed is not increasing.

It says " A bike travels around a circular curve of radius 80m at a constant speed of 10 m/s.
1) Calculate the bike's acceleration.
2) The bike slows uniformly to rewst in 6 seconds. Calculate the tangential acceleration component.
3) The instant the bike is traveling 8 m/s, determine total acceleration.

 

[LowlyPion]

Um, not sure I get all that.
This problem deals with constant speed, so I assume the tangential speed is not increasing.

It says " A bike travels around a circular curve of radius 80m at a constant speed of 10 m/s.
1) Calculate the bike's acceleration.
2) The bike slows uniformly to rest in 6 seconds. Calculate the tangential acceleration component.
3) The instant the bike is traveling 8 m/s, determine total acceleration.

In the first part, you calculate the the V²/r as before.

In the second part the tangential is slowing so there is also a (-) tangential acceleration. This is a vector too.

The third part is asking you when the V is 8, and it's slowing at the negative tangential rate, what is the sum of those 2 vectors. Since the tangential is negative it will be trailing the radial vector at an angle.

 

[veronicak5678]

OK, I think I get it. I'm going to go take a break before I try this. Thanks a lot for helping!