Ferris wheel acceleration physics

[mybrohshi5]

Homework Statement

A mischievous young man riding a ferris wheel decides to release a beanbag as he passes the top point in the ride. The ferris wheel has a radius of 3.39 m and goes around once every 0.95 minutes (or 57 seconds if converted). The bottom of the wheel is 1.03 m off the ground, so that the release point of the bag will be 7.81 m off the ground.

1) What is the direction of the bean bag's acceleration just before it is released by the mischievous young man?

2) What is the speed of the bean bag just before it is released by the mischievous young man?

3) What is the magnitude of the acceleration of the bean bag just before it is released by the mischievous young man?

4) For how much time is the bean bag in the air?

5) The young man finds that the bag does not land directly under the point of release. How far does the bag travel horizontally as it is falling?

Homework Equations

v=2πR/T
a=4π²R/T²

The Attempt at a Solution

1) is the answer up? i thought it was up because he is still rising a little bit until he reaches the very peak of the ferris wheel and releases the bean bag, thus making the direction of the bean bags acceleration up.
the other choices were:
...in the direction of motion of the young man's hand
...the acceleration is zero, thus no direction
...opposite to the direction of motion of the young man's hand
...down

2) v=2π(3.39) / 57 = 0.374 m/s
i wasnt sure about this one because of the extra 1.03 m above the ground and i wasnt sure if this somehow had to be factored in. the answer just seems really slow but i guess ferris wheels don't move fast haha.

3) a=4π²(3.39)/(57)² = 0.0412 m/s²
again not sure about this because I am not sure if i have to factor in the 1.03 m height somehow

4) Not really sure how to find this one...

5) i think i need to know the time before i solve this one...

Thanks for any help!

 

[PhanthomJay]

Homework Statement

A mischievous young man riding a ferris wheel decides to release a beanbag as he passes the top point in the ride. The ferris wheel has a radius of 3.39 m and goes around once every 0.95 minutes (or 57 seconds if converted). The bottom of the wheel is 1.03 m off the ground, so that the release point of the bag will be 7.81 m off the ground.

1) What is the direction of the bean bag's acceleration just before it is released by the mischievous young man?

2) What is the speed of the bean bag just before it is released by the mischievous young man?

3) What is the magnitude of the acceleration of the bean bag just before it is released by the mischievous young man?

4) For how much time is the bean bag in the air?

5) The young man finds that the bag does not land directly under the point of release. How far does the bag travel horizontally as it is falling?

Homework Equations

v=2πR/T
a=4π²R/T²

The Attempt at a Solution

1) is the answer up? i thought it was up because he is still rising a little bit until he reaches the very peak of the ferris wheel and releases the bean bag, thus making the direction of the bean bags acceleration up.
the other choices were:
...in the direction of motion of the young man's hand
...the acceleration is zero, thus no direction
...opposite to the direction of motion of the young man's hand
...down

Incorrect. The beanbag is moving at constant speed, and therefore has no tangential acceleration; however, consider centripetal acceleration...

2) v=2π(3.39) / 57 = 0.374 m/s
i wasnt sure about this one because of the extra 1.03 m above the ground and i wasnt sure if this somehow had to be factored in. the answer just seems really slow but i guess ferris wheels don't move fast haha.

3) a=4π²(3.39)/(57)² = 0.0412 m/s²
again not sure about this because I am not sure if i have to factor in the 1.03 m height somehow

your responses to both 2) and 3) are correct...

4) Not really sure how to find this one...

this is asking how long it is in the air after it is released...use one of the kinematic equations, noting that its initial velocity in the vertical direction is ?

5) i think i need to know the time before i solve this one...

yes, get the time from your answer to 4)..., and then where do you go from there to get the distance??

 

[mybrohshi5]

The acceleration will be towards the center of the ferris wheel so would the answer for 1 be...

1) down? i thought this because with uniform circular motion with constant speed the acceleration is always directed toward the center of the circle which is also known as centripetal acceleration.

2) I am having troubles with this one for finding an equation to use...
so i know the total height = 7.81 m
i know the speed = 0.374 m/s
i know the acceleration = 0.0412 m/s²
i know the acceleration due to gravity = -9.8m/s²

and i need to find time...

do i also need to find the y component of the velocity?

would this be the right equation?
y(t) = y_i + V_yi*(t) + .5at²

i feel like this isn't the correct one to use though cause then it will give me 2 times cause i will have to solve for t using the quadratic equation. can you lead me in the right direction for what equation i should be using?

thank you

ps - do you happen to maybe know a good website that lists all of the one dimension and two dimension kinematic equations? my book seems to do a poor job and giving all the equations and describing how to use them.

 

[PhanthomJay]

The acceleration will be towards the center of the ferris wheel so would the answer for 1 be...

1) down? i thought this because with uniform circular motion with constant speed the acceleration is always directed toward the center of the circle which is also known as centripetal acceleration.

Excellent, my friend.

2) I am having troubles with this one for finding an equation to use...
so i know the total height = 7.81 m
i know the speed = 0.374 m/s
i know the acceleration = 0.0412 m/s²
i know the acceleration due to gravity = -9.8m/s²

and i need to find time...

do i also need to find the y component of the velocity?

yes; the bean bag leaves the top of the wheel with what velocity (magnitude and direction)? This will help you find the initial velocity in the y direction..

would this be the right equation?
y(t) = y_i + V_yi*(t) + .5at²

i feel like this isn't the correct one to use though cause then it will give me 2 times cause i will have to solve for t using the quadratic equation. can you lead me in the right direction for what equation i should be using?

that's more or less the right equation, ...it will simplify when you discover what V_yi is.

 

[mybrohshi5]

ok so i thought about it and i figured that the V_yi is just 0 because it is at the very top of the ride so there is only a horizontal velocity which is 0.347 m/s and there is no y velocity.

if this is correct then here is the work i did:

4) For how much time is the bean bag in the air?

7.81 = 0*(t) + .5(9.8)(t^2)
t = 1.26 seconds

then on part 5:
5) The young man finds that the bag does not land directly under the point of release.
How far does the bag travel horizontally as it is falling?

x(t) = 0 + 0.374(1.26)
x(t) = 0.471 m

are these two correct or am i missing something or was my belief about the V_yi wrong?

thanks :)

 

[mybrohshi5]

Can anyone please check these last two answers in the post above? The assignment is due tomorrow!

Thank you very much for any help.

 

[PhanthomJay]

Can anyone please check these last two answers in the post above? The assignment is due tomorrow!

Thank you very much for any help.

Looks real good!
Nice work.

 

[mybrohshi5]

Thank you so much PhanthomJay. You have been such a great deal of help!

 

[PhanthomJay]

Thank you so much PhanthomJay. You have been such a great deal of help!

Well, it just comes from experience, because, you see, I was that mischievous young man...

 

[mybrohshi5]

Haha!