# Rotational Motion Questions

1. Feb 26, 2008

### bpw91284

1.
Problem Statement
What is the chance of a light car safely rounding an unbanked frictionless curve compared to a heavy car? Both cars have the same speed and tires.
Solution
I know that tan(theta)=v^2/(rg) proves that weight does not matter but I do not understand why and my book doesn't explain it very well. Can some one try explaining this concept to me?

2.
Problem Statement
A rigid massless rod is rotated about one end in a horizontal circle. There is a mass m_1 attached to the center of the rod and a mass m_2 attached to the outer end of the rod. The inner section of the rod sustains three times as much tension as the outer section. Find the ratio m_2/m_1
Solution
Doing a force balance at both m_1 and m_2 shows that the centripetal force (F_c=m*v^2/r) will equal the tension. Though since their velocities will not be the same but their angular velocities will be I replaced velocity with radius*angular velocity.
F_c=m(rw)^2/r

For m_1...
3T=m_1*(rw)^2/r
Solve for T, T=m_1*(rw)^2/(3r)
For m_2...
T=m_2*(2rw)^2/(2r)

Since T=T...
m_1*(rw)^2/(3r)=m_2*(2rw)^2/(2r)
Solving for m_2/m_1 I get 2/3 but my book says the answer is 1/4. Am I wrong or is the book?

2. Feb 26, 2008

### tiny-tim

It's difficult to explain a negative.

Tell us in what way do you think mass would matter, and then we can explain why that's wrong.

The tension in the inner section is the tension from mass m1 plus the tension from mass m2! That should give you the right answer.

(by the way, if you put [noparse]$$before and$$[/noparse] after every equation, it'll print it in LaTeX for you!)

3. Feb 26, 2008

### bpw91284

I understand how there it centripetal force when something is physically attached to say a string and spinning around. I do not understand how the car has centripetal force. To me it just seems like the component of the weight down the slope parallel to the hill is what would hold the car on the track.

4. Feb 26, 2008

### tiny-tim

[size=-2](btw, I think you meant "banked" in your original question - "unbanked" means "flat".)[/size]

It's better not to use the terms "centrifugal force" or "centripetal force".

What matters is the acceleration of the car - if it goes round a circle of radius r at a constant speed v, it has acceleration $$v^2/r$$ towards the centre of the circle.

(It's very important that you fully understand that, and are able to prove it - if not, check it out again in a textbook.)

Both the car and something on a string have this same acceleration.

I think what's bothering you is that, obviously, if the thing-on-a-string is heavier, then the person holding the string has to exert more force (force = mass times acceleration), and therefore the weight of the thing-on-a-string is extremely important.

But with the car, although more force is needed, it's effectively being supplied by the car itself - it's the force of gravity times the weight of the car!

So the weight of the car doesn't matter!

5. Feb 26, 2008

### bpw91284

That does not make sense according to the problem because then tension at mass one would be 4 times the tension at mass two, and the problem states that it's three times. Also, I did it that way and still did not get the correct answer. If you could show me the math to get m_2/m_1=1/4 I'd really appreciate it.

For m_1...
$$4T=m_1*(rw)^2/r$$
For m_2...
$$T=m_2*(2rw)^2/(2r)$$

I get m_2/m_1=0.125

6. Feb 26, 2008

### tiny-tim

tension is force.

force = mass x acceleration = mass x $$w^2$$ x r

$$w^2$$ is the same for both.

So 1.m1 + 2.m2 = 3(2.m2)

So m1 = 2(2.m2) = 4.m2.

7. Feb 26, 2008

### bpw91284

What are those equations from?

8. Feb 26, 2008

### tiny-tim

$$r_1.m_1.w^2 + r_2.m_2.w^2 = 3(r_2.m_2.w^2)$$​

Divide by $$w^2$$, and put $$r_1\,=\,1\qquad r_2\,=\,2\,:$$

$$\,m_1 + 2.m_2 = 3(2.m_2)\,.$$

9. Feb 26, 2008

### bpw91284

Where are you doing sum of forces = ma at? And why doesn't my method work?

10. Feb 27, 2008

### tiny-tim

It's not "where", it's "which direction" - I'm doing it in the direction along the rod.

Because, for some reason, you've put an extra 4 in.

(But I don't understand how you got .125 instead of .0625 - did you fail to cancel one of the 2s?)

11. Feb 27, 2008

### bpw91284

Alright, reread the problem and tell me what to change in my method.
Problem Statement
A rigid massless rod is rotated about one end in a horizontal circle. There is a mass m_1 attached to the center of the rod and a mass m_2 attached to the outer end of the rod. The inner section of the rod sustains three times as much tension as the outer section. Find the ratio m_2/m_1

Solution
For m_1...
$$3T=m_1*(rw)^2/r$$
Solve for T
$$T=m_1*(rw)^2/(3r)$$
For m_2...
$$T=m_2*(2rw)^2/(2r)$$

Since T=T...
$$m_1*(rw)^2/(3r)=m_2*(2rw)^2/(2r)$$
Solving for m_2/m_1 I get 2/3 but my book says the answer is 1/4.

12. Feb 27, 2008

### tiny-tim

(I see you've corrected the 4 in the first line to a 3.)

Actually, with your formula, you don't get 2/3, you get 1/2.3 (= 1/6):
$$m_1*(rw)^2/(3r)=m_2*(2rw)^2/(2r)$$​

But that formula is wrong anyway, because, as I said before:
So it should be:
$$m_1*(rw)^2/(r)\,+\,m_2*(2rw)^2/(2r)\,=\,3.m_2*(2rw)^2/(2r)$$​
which is the same as my formula:
and they both give 1/4, as in your book.

13. Feb 27, 2008

### blochwave

Right, to comment, what YOU(topic creator) solved was the problem "You have two masses on two separate massless rods, and one rod is half the length of the other and they have the same angular velocity..."

There is tension on the rod because of m2, and it affects the WHOLE rod, inner and outer. There is tension because of m1, and it affects only the inner. Two things contribute to the tension in the inner rod

14. Feb 27, 2008

### bpw91284

Wow, thanks I finally get it. The thing I was misunderstanding is what they meant by "sections". Thanks for all the help.