Rotational Motion and magnitude

[artsakh]

Im studying for a final, there are a few questions about rotational motion that i have, can anyone please help?

A 5.2 kg disk of radius 2.3 m initially rotates about its axis at 6.4 rad/s. A tangential 7.4-N
frictional force is applied to the rim of the disk. How many revolutions does the disk make
before coming to stop?

A circular bird feeder 20 cm in radius has a moment of inertia of 0.12 kg⋅m2. The feeder is
suspended by a thin wire and is spinning slowly at 0.60 rad/s. A 150 g bird lands on the rim of
the feeder, coming in tangent to the rim at 1.5 m/s in a direction opposite the feeder's rotation.
What is the rotation rate after the bird lands?

Find the magnitude of the angular momentum of a 70 g particle with respect to the origin
when the particle is at x = 5 m, y = -3 m, and its velocity is given by 6i + 3j m/s.

I really don't know where even to begin with these questions, especially the last one. Any help would be greatly appreciated

 

[Fewmet]

Im studying for a final, there are a few questions about rotational motion that i have, can anyone please help?

A 5.2 kg disk of radius 2.3 m initially rotates about its axis at 6.4 rad/s. A tangential 7.4-N
frictional force is applied to the rim of the disk. How many revolutions does the disk make
before coming to stop?

Could you answer a question about how far a car of given mass and initial velocity goes before stopping under the effects of a specified force? That's analogous this.

You can easily find the torque from what is given, and then the rotational acceleration...does that let you see the path to the answer?

 

[Artius]

This wouldn't happen to be NYU-Poly 2010 final would it? I was about to make a post on one of the questions lisetd here myself, and all of them are on the final from last year

 

[artsakh]

Could you answer a question about how far a car of given mass and initial velocity goes before stopping under the effects of a specified force? That's analogous this.

You can easily find the torque from what is given, and then the rotational acceleration...does that let you see the path to the answer?

I have the formula $$I_{disk}$$=1/2MR^2, and T=I$$\alpha$$, and also $$a_{r}$$=$$\omega$$^2r that'll give me the I and the rotational acceleration? i still don't know what to do after though...

 

[artsakh]

This wouldn't happen to be NYU-Poly 2010 final would it? I was about to make a post on one of the questions lisetd here myself, and all of them are on the final from last year

yea it is

 

[Fewmet]

I have the formula $$I_{disk}$$=1/2MR^2, and T=I$$\alpha$$, and also $$a_{r}$$=$$\omega$$^2r that'll give me the I and the rotational acceleration? i still don't know what to do after though...

Do you have any other equations to work with that involve $$\alpha$$?

 

[artsakh]

Do you have any other equations to work with that involve $$\alpha$$?

There's many, here is the formula sheet we are provided with:
http://i52.tinypic.com/5k34ic.jpg

 

[Fewmet]

Good.

So you want to find how many rotations the disk disk goes through before stopping. In others words, you want the angle it goes though (which has the symbol $$\theta$$).

In that list of equations is one that has $$\theta$$, angular acceleration and other variables for which you have values. You just need to identify it, substitute and solve.

Can you do it from there?

 

[Artius]

I'm going to take a shot at solving the first problem, if not, I think I'll at least set it up


Since it's a rigid body: Rotational inertia = Sigma(mi ri^2)

therefore I = 5.2 * 2.3^2 = 5.2 * 5.29 = 27.51 kg * m^2

Since I serves as a sort of replacement in rotation for "mass" in translational motion (correct me if I'm mistaken), we have three Values to work with right now

OmegaInitial = +6.4 rad/s

I = 27.51 kg * m^2

Frictional Force = 7.4 N

Now let's see...how to relate this to angular acceleration?

We have I * alpha = Torque...
So Torque / Moment of Inertia = alpha

Torque = Radius * Force
Radius = 2.3
Force = -7.4 N
Torque = -17.04 N*m

-17.04 N*m / 27.51 = -0.62 rad/s^2 (? wait...the units don't add up)

Going to try to put this into the calculator anyway... Wf = Wi + at
0 = 6.4 - 0.62t
t = 10.3 seconds

ChangeinTheta = 10.3 * (6.4) - 0.31(10.3)^2

ChangeInTheta = 33.0321 radians
Revolutions = 5.26 Revolutions

...where did i make a mistake? none of the answers match with mine =/

 

[Fewmet]

Since it's a rigid body: Rotational inertia = Sigma(mi ri^2)

therefore I = 5.2 * 2.3^2 = 5.

I for a disk is 1/2 Mr².

Your general approach looks correct. You need not find time: there is a relationship involving what you are given that doesn't use time. But you way is fine.

Can you correct the the I and see if that gives you one of the answer choices? If not, I'll look more carefully to see if there is a problem.

(Sorry about not being thorough myself, but I am writing up an exam for my students).

 

[artsakh]

yea, the problem was with the I, the correct answer is 2.63

 

[Artius]

2 Questions of my own involving Gravity.

If you have two moons orbiting the same planet in circular paths where Moon A has an orbital radius 2.4 greater than Moon B, how do you find the ratio of their orbital speeds?

and another question from the same exam

A moon of mass 3.2 * 10^22 orbits a planet. The distance of the moon's center from the center is 27 Gigameters. If the center of mass of the moon-planet system is 2 Gm from the planet's center, what is the mass of the planet? I'm guessing a setup that looks like:

0 = Position of Planet

CoM = 2

2Gm = 0 + (32e22)(27Gm) / (32e22 + M1) -> solve for M1?

 

[Fewmet]

yea, the problem was with the I, the correct answer is 2.63

Great. Thanks for letting me know.

artsakh: I took a look at the second one (the one about the bird). It's just about the conservation of rotational momentum. It is probably counterintuitive, but the rotational speed of the bird is $$\frac{v}{r}$$. Do you see where it goes from there?

 

[rock.freak667]

Another way to solve the first question is to use conservation of energy.

rotational ke of disk = work done by friction

work done by friction = torque*angular displacement

2nd:

Think momentum conservation for this.

3rd: How is angular momentum defined in Cartesian coordinates? (in terms of cross product)

 

[Fewmet]

Another way to solve the first question is to use conservation of energy.

rotational ke of disk = work done by friction

work done by friction = torque*angular displacement

That's very pretty. I wish I'd seen it that way at first.

 

[rock.freak667]

2 Questions of my own involving Gravity.

If you have two moons orbiting the same planet in circular paths where Moon A has an orbital radius 2.4 greater than Moon B, how do you find the ratio of their orbital speeds?

Use the fact that the centripetal force = gravitational force for each moon, then make v² the subject, then divide.?[/QUOTE]

That's very pretty. I wish I'd seen it that way at first.

Always look for the simplest way

 

[Fewmet]

A moon of mass 3.2 * 10^22 orbits a planet. The distance of the moon's center from the center is 27 Gigameters. If the center of mass of the moon-planet system is 2 Gm from the planet's center, what is the mass of the planet? I'm guessing a setup that looks like:

0 = Position of Planet

CoM = 2

2Gm = 0 + (32e22)(27Gm) / (32e22 + M1) -> solve for M1?

I don't see where that comes from.

Do you know that the product of each mass from its distance to the center of mass is the same?

 

[artsakh]

Great. Thanks for letting me know.

artsakh: I took a look at the second one (the one about the bird). It's just about the conservation of rotational momentum. It is probably counterintuitive, but the rotational speed of the bird is $$\frac{v}{r}$$. Do you see where it goes from there?

Not really, I am sorry for being such a pain though :shy:

The weight of the feeder can be calculated because r and I are known, and it is 6kg, the $$\omega$$ of the bird comes out to be -7.5 (but what are the units? seconds?) i don't really see where to go from here...

 

[Artius]

I don't see where that comes from.

Do you know that the product of each mass from its distance to the center of mass is the same?

So Xcom = ∑ (1, n) mx / ∑ (1, n) m doesn't work here?

 

[Fewmet]

So Xcom = ∑ (1, n) mx / ∑ (1, n) m doesn't work here?

Sorry: I was framing it differently and didn't recognize where you were starting. Yes: that works (and is mathematically equivalent to what I was suggesting). I was just putting the center of mass at zero Gm.

 

[Artius]

Use the fact that the centripetal force = gravitational force for each moon, then make v² the subject, then divide.?

Always look for the simplest way [/QUOTE]

As in
mv^2 /r = GMm / r^2,
m2v2^2 / 2.7r = GMm2 / (2.7r)^2?

I wish a certain tenured professor would see things your way

 

[Fewmet]

Not really, I am sorry for being such a pain though :shy:

The weight of the feeder can be calculated because r and I are known, and it is 6kg, the $$\omega$$ of the bird comes out to be -7.5 (but what are the units? seconds?) i don't really see where to go from here...

You are not being a pain. It is tough to lead you through this because I don't know how the material was framed for you or how you interpret it. It is clever that you tried to get mass from the rotational inertia. In a rotational problem, though, the rotational inertia is more useful.

Can you find the rotational momentum of the feeder? When the bird lands, it adds it rotational momentum to feeder. From there you can find the rotational speed after it lands.

 

[artsakh]

You are not being a pain. It is tough to lead you through this because I don't know how the material was framed for you or how you interpret it. It is clever that you tried to get mass from the rotational inertia. In a rotational problem, though, the rotational inertia is more useful.

Can you find the rotational momentum of the feeder? When the bird lands, it adds it rotational momentum to feeder. From there you can find the rotational speed after it lands.

Well L=r$$\times$$p=I$$\omega$$, so the L of the feeder is 0.12*0.6? I don't understand what you mean by the bird adding its rotational momentum. The momentum of the bird is .150*1.5 tangent and opposite to the rotation of the feeder isn't it? You can get the $$\omega$$ of the bird but i don't see how that fits in...

 

[rock.freak667]

Always look for the simplest way

As in
mv^2 /r = GMm / r^2,
m2v2^2 / 2.7r = GMm2 / (2.7r)^2?

I wish a certain tenured professor would see things your way [/QUOTE]

That would give you the velocity of Moon B.

So the same with A and you will see that you will get

v_B² = k*GM/r

v_A²= GM/2

and just divide and take the square root to get the ratio you need.

 

[Rayquesto]

A 5.2 kg disk of radius 2.3 m initially rotates about its axis at 6.4 rad/s. A tangential 7.4-N
frictional force is applied to the rim of the disk. How many revolutions does the disk make
before coming to stop?

Ok! so, first thing is you got to find how much distance the disk circulates before coming to rest. I'm assuming that due to the fact that normal and gravitational force cancel each other out, the only force that circular is the 7.4N frictional force. Assuming this, it is essential to know the disks acceleration, but let's just write out the displacement formula first. So, displacement=omegainitial times time minus 1/2 of acceleration times (time)^2. Ok, next we find acceleration. 7.4=mass times a=mass times (tangential speed)^2/radius. Acceleration is 1.42m/s^2. rad/sec is not the same as meters/sec. So, in order to find meters per second, make 1.42m/s^2=mass times tangential speed^2/radius. so, 1.42m/s^2= tangential speed^2/2.3meters. So, tangential speed=1.8m/s=omega. so next thing we do is find the amount of time it took to stop. assuming this tangential speed is constant, omega=acceleration times time. so, (1.42m/s^2)Times time=1.8m/s and time=1.27seconds. next thing is calculate displacement. SO, displacement=(1.8m/s)(1.27seconds) plus 1/2(1.42m/s^2)(1.27s^2)=3.43meters and to find the number of revs you say #revs=3.43meters/circumference=3.43meters/2pie2.3meters=only .24 or 1/4 of its circumference. I could be wrong but imagining this application is like watching a spinning top spin due to some kind of string that had barely any force. I could be horrifyingly wrong. Please anyone explain to me my wrongness! lol please?

 

[Fewmet]

artsakh: Thinking back though this problem, I see I am unsure of one point. It's been a long time since I've worked one like this. I think that the rotational speed of the bird is
$$\omega$$=$$\frac{v}{r}$$=$$\frac{.15 m/s}{.2 m}$$ = 0.75/s. And I think its moment of inertia is mr². That let's you find its rotational momentum.

Something feels wrong about this and (having awoken pre-dawn to see the "www.youtube.com/watch?v=__RLPmenKeo"[/URL]), I'm not thinking clearly enough at the moment to see how to derive it.

I am sorry I cannot be of more help. Maybe rock.freak667 or someone else can take you the next step. I have to get some sleep before another attempt to find all four planets...

 

[Rayquesto]

A 5.2 kg disk of radius 2.3 m initially rotates about its axis at 6.4 rad/s. A tangential 7.4-N
frictional force is applied to the rim of the disk. How many revolutions does the disk make
before coming to stop?

Ok! so, first thing is you got to find how much distance the disk circulates before coming to rest. I'm assuming that due to the fact that normal and gravitational force cancel each other out, the only force that circular is the 7.4N frictional force. Assuming this, it is essential to know the disks acceleration, but let's just write out the displacement formula first. So, displacement=omegainitial times time minus 1/2 of acceleration times (time)^2. Ok, next we find acceleration. 7.4=mass times a=mass times (tangential speed)^2/radius. Acceleration is 1.42m/s^2. rad/sec is not the same as meters/sec. So, in order to find meters per second, make 1.42m/s^2=mass times tangential speed^2/radius. so, 1.42m/s^2= tangential speed^2/2.3meters. So, tangential speed=1.8m/s=omega. so next thing we do is find the amount of time it took to stop. assuming this tangential speed is constant, omega=acceleration times time. so, (1.42m/s^2)Times time=1.8m/s and time=1.27seconds. next thing is calculate displacement. SO, displacement=(1.8m/s)(1.27seconds) plus 1/2(1.42m/s^2)(1.27s^2)=3.43meters and to find the number of revs you say #revs=3.43meters/circumference=3.43meters/2pie2.3meters=only .24 or 1/4 of its circumference. I could be wrong but imagining this application is like watching a spinning top spin due to some kind of string that had barely any force. I could be horrifyingly wrong. Please anyone explain to me my wrongness! lol please?

ok I am correcting my mistake. yea that answer i put didnt seem right. you don't find the tangential speed. you find the angular speed. there's a difference. so instead use 6.4rad/second multiply it times 2pie to get 40meters/second then plug this into the equation to get the amount of meters the disk circulated if it went at an accerelation of 1.42m/s^2. so, displacement=40m/s times 1.27 seconds plus 1/2 times 1.42m/s^2 times 1.27s^2=52meters. then (52meters)/(2pie times 2.3 meters)=3.6 revolutions i could still be wrong but correct me please. I am in regular high school physics. so, I am no expert but id like to be one of these days.

 

[artsakh]

A 5.2 kg disk of radius 2.3 m initially rotates about its axis at 6.4 rad/s. A tangential 7.4-N
frictional force is applied to the rim of the disk. How many revolutions does the disk make
before coming to stop?

Ok! so, first thing is you got to find how much distance the disk circulates before coming to rest. I'm assuming that due to the fact that normal and gravitational force cancel each other out, the only force that circular is the 7.4N frictional force. Assuming this, it is essential to know the disks acceleration, but let's just write out the displacement formula first. So, displacement=omegainitial times time minus 1/2 of acceleration times (time)^2. Ok, next we find acceleration. 7.4=mass times a=mass times (tangential speed)^2/radius. Acceleration is 1.42m/s^2. rad/sec is not the same as meters/sec. So, in order to find meters per second, make 1.42m/s^2=mass times tangential speed^2/radius. so, 1.42m/s^2= tangential speed^2/2.3meters. So, tangential speed=1.8m/s=omega. so next thing we do is find the amount of time it took to stop. assuming this tangential speed is constant, omega=acceleration times time. so, (1.42m/s^2)Times time=1.8m/s and time=1.27seconds. next thing is calculate displacement. SO, displacement=(1.8m/s)(1.27seconds) plus 1/2(1.42m/s^2)(1.27s^2)=3.43meters and to find the number of revs you say #revs=3.43meters/circumference=3.43meters/2pie2.3meters=only .24 or 1/4 of its circumference. I could be wrong but imagining this application is like watching a spinning top spin due to some kind of string that had barely any force. I could be horrifyingly wrong. Please anyone explain to me my wrongness! lol please?

ok I am correcting my mistake. yea that answer i put didnt seem right. you don't find the tangential speed. you find the angular speed. there's a difference. so instead use 6.4rad/second multiply it times 2pie to get 40meters/second then plug this into the equation to get the amount of meters the disk circulated if it went at an accerelation of 1.42m/s^2. so, displacement=40m/s times 1.27 seconds plus 1/2 times 1.42m/s^2 times 1.27s^2=52meters. then (52meters)/(2pie times 2.3 meters)=3.6 revolutions i could still be wrong but correct me please. I am in regular high school physics. so, I am no expert but id like to be one of these days.

Thank you very much but we already figured this one out, its the other two that i need help with

 

[lepton5]

Thank you very much but we already figured this one out, its the other two that i need help with

for your last two problem, can you solved it based on advice from rock.freak667 ?

2nd:

Think momentum conservation for this.

3rd: How is angular momentum defined in Cartesian coordinates? (in terms of cross product)

edit: for 2nd problem ~ use angular momentum conservation