# Rotational Motion and Gravity/Planetary Orbits review

1. Jul 26, 2008

### vdfortd

I have a test Monday and i'm trying some review problems my professor gave us to study from. I'm not looking for answers i'm just looking for someone to help step me through a few of these problems. Thanks for any and all help you guys can provide.

1. The problem statement, all variables and given/known data
3a). After starting, a lawn mower blade turns at a constant angular velocity of 20 rad/s. In a 10s interval, how many radians and how many degrees does it turn?
----------
I figured out this part. It's 200 radians (20X10) and 11459.156 degrees ( 200(180/$$\pi$$))

b) If the blade has a radius of .3 m, what is the linear speed of the tip of the blade?
------
Using v=rω, I got 6 by doing (.3)(20)= 6

c) If the blade is considered as a rectangular plate of mass .4 kg, and a width of .05 m, what is its kinetic energy as it spins.

This question (3c) is what I am stuck on.

2. Relevant equations

Moment of Inertia for a rectangular plate is: I= 1/12M(a2+b2) where a and b are the length and width of the rectangle.

Kinetic Energy is : K=(1/2)Iω2

Edit: I dont don't why the omega symbol is there as a superscript. It should be omega squared. Sorry i'm trying to get used to this.

3. The attempt at a solution

Since I know the equations, i was going to plug the mass and dimensions into the equation I= 1/12M(a2+b2) to get Inertia but I realized that they only gave me one dimension (the width) and I don't know what to do since I was not given a second dimension. I'm not sure if my teacher forgot to include the second dimension or if this is actually part of the problem.

After i found Inertia i was going to use the equation K=(1/2)Iω2 since I knew what the Inertia and what Omega is, and use that to find the Kinetic Engery. I'm just at a standstill with the first equation.

1. The problem statement, all variables and given/known data
6. A particle of mass .4kg is attached to the 100-cm mark of a meter stick of mass .1 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 4.00 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the 50cm mark and (b) perpendicular to the table through the 0 cm mark.

mass of particle = .4kg
mass of meter stick =.1kg
Lengh of stick = 100cm = 1 m

2. Relevant equations

For part a) Moment of Inertia for a Long thin rod with rotation through center: I= (1/12)ML2

For part b) Moment of Inertia for a Long thin rod with rotation through end: I= (1/3)ML2

For these equations, L = the length of the rod and M = the total mass

To find angular momentum, use the equation L= Iω
L represents angular momentum in this equation.

3. The attempt at a solution

I know the answers. a) is .433 kgm2/s and b) is 1.73 kgm2/s.

But when I do the problem I don't get the right answers.

For (a), I did I= (1/12)(.5)(12)

where .5 is the total mass of the stick and the particle and 1 is the length of the stick in meters. I got about .04167 and when I mutiplied that by omega (which is 4) I got .16667

Last edited: Jul 26, 2008
2. Jul 26, 2008

### cryptoguy

well for your first question, you're given a radius in part B.

edit:
for second problem you forgot to include the particle in calculating inertia

Last edited: Jul 26, 2008
3. Jul 26, 2008

### tiny-tim

Welcome to PF!

Hi vdfortd ! Welcome to PF!
Isn't that for an axis perpendicular to the rectangle?

Nooo … you can't add the masses.

The MI formula for a rod is for a uniform rod.

You must calculate the two MIs separately, and then add them.

4. Jul 26, 2008

### vdfortd

Yeah, I brought the problems with me to work and read through it again and realized that they gave the radius in part B. I feel stupid for asking that now.

I realized what I did wrong in the first problem. I missed the fact that they gave me a radius in part B of the question.

As for the second problem, I didn't know you had to add them seperatedly. I'm going to go try that. Thanks.

5. Jul 27, 2008

### vdfortd

Re: Welcome to PF!

OK, I thought I understood what you meant but i got the same answer :(. I did the first part for the .4kg (I= (1/12)ML2. So I did I= (1/12)(.4)(12)= .03333. Then I did it for the mass of the meter stick. I=(1/12)(.1)(12) and I got .0083333. Add them together and you get .0416 which is the same number I got when I added the masses together (.5kg) and plugged it into the equation.

Then for L= Iω .. I got L= 4(.0416) = .166533. The correct answer is .433.

If I work this problem backwards and do .433= I(4), you get that I= .10825. So I have to somehow get .10825 from the first equation and I cant see what I'm doing wrong.

6. Jul 27, 2008

### cryptoguy

Don't calculate the inertia of a .4 kg particle using the formula for a long thin rod. Imagine that just the particle was revolving (held by a massless string for example). What would its' inertia be? Add this inertia to the inertia of the meter stick.

7. Jul 27, 2008

### tiny-tim

Hi vdfortd!

1/12 etc is only for a rod, about an axis through its centre.

The .4kg is "a particle" (which means the examiner is telling you to treat it as a point), and about a distant axis.

So its MI about that axis is … ?

8. Jul 27, 2008

### vdfortd

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.

So I do (1/12)(.1)(1) for the Inertia of the meter stick and then I find the Inertia of the particle itself which I assume would be the equation I=mr2. The radius would be how far the particle is from the center of the meter stick, which is 50 cm or .5 meters. So the Inertia of the particle would be I= (.4)(.52). Does that sound right?

9. Jul 27, 2008

### tiny-tim

That's it!

(except of course, "inertia" is mass … so don't leave out "moment of" )

10. Jul 27, 2008

### vdfortd

I just tried both part a and b and they worked out to the correct answer. Thanks a lot. My professor never really explained how to do these type or problems, he just told us the equations for certain objects are listed in the book. Anyway, I finished my review so thanks again.