Merry Go-Round Circular Motion

In summary, the man on the carnival merry-go-round has a constant speed of 3.66 m/s and a centripetal acceleration of 1.83 m/s2. In order to find the position vector, we can use the formula a = v^2/r and T = 2pi r/v. Part (a) asks for the magnitude of the position vector, which is equivalent to the radius in this situation. Parts (b) and (c) ask for the direction of the position vector when the acceleration is directed due east and due south, respectively. For part (a), we can solve for r by rearranging the formula and plugging in the values for v and a. For parts (b) and (
  • #1
Shatzkinator
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Homework Statement


A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.66 m/s and a centripetal acceleration a of magnitude 1.83 m/s2. Position vector r locates him relative to the rotation axis. a) what is the magnitude of r? What is the direction of r when a is directed b) due east and c) due south?


Homework Equations


a = v^2/r
T = 2pi r/v



The Attempt at a Solution


a) no idea how to do this part.
b) i thought since acceleration is east, which would be toward centre because circular motion, then wouldn't position be same direction??
c) same as b
 
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  • #2
You have a = v^2/r and T = 2pi r/v.
Looking for r. Know v and a.
Which of the formulas has r, v and a in it? Solve it for r, then put in the values for v and a.
 
  • #3
Delphi51 said:
You have a = v^2/r and T = 2pi r/v.
Looking for r. Know v and a.
Which of the formulas has r, v and a in it? Solve it for r, then put in the values for v and a.

Hi I wish it was that easy. But they're looking for the position vector, not the radius, there should be a little arrow over the r that I don't know how to do. =)

sorry for the misunderstanding.
 
  • #4
Part (a) asks for the magnitude of r .
Okay, on part (b): What is the direction of r when a is directed b) due east
sketch the thing and picture the man going around in circles with an acceleration vector on himself. Stop when his arrow goes east. What is the direction of the radius to/from the man at that point? Hmm, does r go toward the center or away from the center? Hope you know - I don't.
 
  • #5
Delphi51 said:
Part (a) asks for the magnitude of r .
Okay, on part (b): What is the direction of r when a is directed b) due east
sketch the thing and picture the man going around in circles with an acceleration vector on himself. Stop when his arrow goes east. What is the direction of the radius to/from the man at that point? Hmm, does r go toward the center or away from the center? Hope you know - I don't.

Okay i understand b and c now.

But a) is asking for the magnitude of the position vector? Not the radius because radius is just r, not r with an arrow above it. =S
 
  • #6
"Position vector relative to the axis" . . . sounds like radius to me!
Ah, but it answers my question - you want the direction from the axis to the man in parts b and c.
 
  • #7
Delphi51 said:
"Position vector relative to the axis" . . . sounds like radius to me!
Ah, but it answers my question - you want the direction from the axis to the man in parts b and c.

wow I'm an idiot.

Thank you =)
 
  • #8
Another problem:

A purse at radius 2.00 m and a wallet at radius 3.00 travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2)i + (4.00 m/s2)j, at that instant and in unit-vector notation, what is the acceleration of the wallet?

equation:

a = v^2/r

attempt:

i tried substituting v = square root (3a) into the equation a = v^2/r for the purse, because they both have same velocity? Uniform circular motion? Not working...
 
  • #9
I would use the other acceleration formula for this one, with v replaced by 2(pi)r/T.
Knowing the magnitude of acceleration at r = 2, you can find the T. Then use the formula again to get the magnitude at 3. Or just look at the formula and ask yourself what increasing r by a factor of 1.5 does to acceleration.

The direction of the acceleration will be the same for the wallet as it is for the purse.
 
  • #10
Delphi51 said:
I would use the other acceleration formula for this one, with v replaced by 2(pi)r/T.
Knowing the magnitude of acceleration at r = 2, you can find the T. Then use the formula again to get the magnitude at 3. Or just look at the formula and ask yourself what increasing r by a factor of 1.5 does to acceleration.

The direction of the acceleration will be the same for the wallet as it is for the purse.

the answer should be (3m/s^2 i + 6m/s^2 j).. Now I did what you said but I'm not sure when calculating T how to keep everything in unit vector notation.. it all seems to be really weird.
 
  • #11
a = 4(pi)^2*m*r/T^2 so multiplying r by 1.5 makes acceleration 1.5 times bigger.
The direction doesn't change, so both the x and y components must be multiplied by 1.5.
So (2, 4) goes to (3, 6).
 

Related to Merry Go-Round Circular Motion

1. What is circular motion?

Circular motion is the movement of an object along a circular path, where the object maintains a constant distance from a fixed point known as the center of the circle.

2. How does a merry-go-round exhibit circular motion?

A merry-go-round exhibits circular motion because it rotates around a central axis, with objects on the outer edge following a circular path.

3. What is the relationship between linear and circular motion?

Linear motion is a type of motion where an object moves along a straight line, while circular motion is a type of motion where an object moves along a circular path. Linear motion can be considered as a special case of circular motion, where the radius of the circle is infinitely large.

4. How does the speed of an object on a merry-go-round change?

The speed of an object on a merry-go-round changes as the object moves closer to or further away from the center of the circle. This is because the distance from the center affects the length of the circular path, therefore changing the speed at which the object moves.

5. What is the role of centripetal force in circular motion?

Centripetal force is the force that acts towards the center of the circle, keeping an object in circular motion. It is necessary to counteract the natural tendency of an object to move in a straight line and ensure that it follows the circular path.

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