What Is the Magnitude and Direction of the Position Vector on a Merry-Go-Round?

[Shatzkinator]

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

 

[Delphi51]

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.

 

[Shatzkinator]

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.

 

[Delphi51]

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.

 

[Shatzkinator]

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

 

[Delphi51]

"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.

 

[Shatzkinator]

"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 =)

 

[Shatzkinator]

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...

 

[Delphi51]

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.

 

[Shatzkinator]

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.

 

[Delphi51]

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).