In Figure 10-30, wheel A of radius rA = 15 cm is coupled by belt B to wheel C of radius rC = 22 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s2. Find the time needed for wheel C to reach an angular speed of 110 rev/min, assuming the belt does not slip.



I tried by setting up a ratio to find that it takes 75rev/min with wheel A. Afterwards, I found the velocity by multiplying 75 rev/min by (2 * pi * .15 m) then divide that number by 60s. This net me the angular speed of 1.18 rad/s.
Next, I found the time it takes by using the equation angular speed = angular speed initial + angular acceleration * time. I end up with .736 s as a result.
Am I on the right track? If so, I'm not quite sure what I should do next. I though it would involve another ratio, but so far that is not working for me.

Just use a calculator.

working on it in my luch break....

Method I think I'm going for (results pending)

find angular speed of small wheel to drive large wheel at 110rev/min, then work back to figure out how long it will take to get the small wheel up to that speed if it accellerates constantly at 1.6 rad/s^2...


...

ok, for every rev of the small wheel, you feed not quite enough drive to perform one rev of the large wheel - the factor being 15:22.

So, take 22/15 (= 1.46) x 110rev/min = 161.333 rev min needed in the small wheel to drive the large wheel at 110.

1 rev/min = 2xpi/60 rad/s = 0.1047

0.1047 x 161.333 = 16.89 rad/sec needed.

then it's just using v = u + at to arrive at 10.55s to achieve the desired velocity.

Hope that's right.

(and Booyah if it is)

1 revolution is basically the circumference (2*pi*r) of the circle right? If that is the case, should it be (2 * pi * 0.15 m) / 60 rad/s ?

Its quite a simple problem, really. the two wheels are coupled by a chain that does not slip. that means that the tangential velocitya nd acceleration of each gear are exactly the same. therefore, using the equation

At= (angular acceleration)R , where At is the tangential acceleration,

you can equate the two together for each wheel.

Ra = Radius of gear A and Rc = Radius of Gear C, so

(angular acceleration of gear A)Ra=(angular acceleration of gear C)Rc

finding the angular acceleration of Gear C, you can then use the simple equation

(Final Angular speed)/(Angular Acceleration) = t

You have to change the final speed of wheel C from Rev/min to Rad/s, but that's a very simple conversion that I'm going to assume you already did.

Results pending

Thisn is the answer, i put it in spoiler tags in cae you didn't want to be given it

The answer is 10.559s

Edit:



Uh, yeah, that's correct for the first half (that circumference = 2piR, but circumfrence /= radiens. In every circle tehre are 2pi radiens, no matter the radius, so 1 rev = 2pi radiens, hense his equation that 1rev/min= 2pi/60sec

Oh okay, thank you very much guys.

/me walks in and see's Martyn's and Noble's answers...

/me nod's and walks out.

Interesting question btw, never come across it before.

I've done it a quicker way when I was studying gears in my 3rd year, but I decomposed the method to remove all the shortcuts - I think it's much more obvious to show all the steps.

Oh, and it was simpler than I thought it would be - I was expecting to fall on my face and get all embarrassed!

How do you guys get so good at this? I'm having a terrible time handling physics. It's bad enough that my professor is quite hard to understand due to his accent, but he also spends alot of time deriving equations instead of explaining the concepts more indepth and applying them to problems.

This much is simple knowledge of kinematics and geometry, nothing too bad. Only a few basic principles you need to know to be able to do this stuff

In Figure 10-42 a cylinder having a mass of 3.0 kg can rotate about its central axis through point O. Forces are applied as shown: F1 = 12.0 N, F2 = 8.0 N, F3 = 4.0 N, and F4 = 5.0 N. Also, r = 5.0 cm and R = 12 cm.



Find the magnitude and direction of the angular acceleration of the cylinder. (During the rotation, the forces maintain their same angles relative to the cylinder.)
Magnitude

Okay....so why doesn't T(torque) = mR^2a(angular acceleration) work?
Isn't it just a = (.05m *12N)+(.05m*8.0N)+(.05m*4.0N) / ((3.0kg)(.12m))^2





Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released.



a.What is the magnitude of the initial acceleration of particle 1?
b.What is the magnitude of the initial acceleration of particle 2?

I have no idea where to begin with this problem when the only thing they give me is length.





In Figure 10-53, two blocks, of mass m1 = 310 g and m2 = 630 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.



Find the magnitude of the acceleration of the blocks.
a= 2.635 m/s^2 (I got the right answer, but I don't specifically understand the steps to getting it)
b. Find the tension of T1
c. Find the tensin of T2

You have to take into account that some of those forces are acting in different directions and so you ahve to assign one direction positive (i.e. clockwise), and then forces going the otehr way (coutnerclockwise) are negative. Your looking for the net torque, so have

Tnet=(F2*R)+(F3*r)-(F1*R)=(8*.12)+(4*.05)-(12*.12)= -.28 (or .28 counterclockwise)

Then

Tnet=mR^2(angluar acceleration), so, if we take counterclockwise as positive,

.28=3*(.12^2)(angular acceleration) so

Angular acceleration = .28/(3*(.12^2)) = 6.48 Rad/s^2 counterclockwise.




You don;t say what youre solving for... give me that and i can explain it, it looks very simple.



Sorry, i'm tired, i'll leave this for someone else. nd i was never great with pullys anyway (well, after they stopped being considered massless)

Use free body diagrams to analyse the last one - and yeah, give us the full problem for the other one - it looks like a simple moment system (or maybe some strain energy thing too).

I don't have enough battery on my laptop to do any more than this until i get t'internet at my house, sorry!

[QUOTE=Noble Outlaw]You have to take into account that some of those forces are acting in different directions and so you ahve to assign one direction positive (i.e. clockwise), and then forces going the otehr way (coutnerclockwise) are negative. Your looking for the net torque, so have

Tnet=(F2*R)+(F3*r)-(F1*R)=(8*.12)+(4*.05)-(12*.12)= -.28 (or .28 counterclockwise)

Then

Tnet=mR^2(angluar acceleration), so, if we take counterclockwise as positive,

.28=3*(.12^2)(angular acceleration) so

Angular acceleration = .28/(3*(.12^2)) = 6.48 Rad/s^2 counterclockwise.
[/QUOTE]


That actually isn't the correct answer. However, I was eventually able to kind of figure it out.
Tnet = Ia(angular acceleration)
(.12m * 12N) - (.12m * 8.0N) - (.05m * 4N) = (1/2 * 3.0 kg * (.12m) ^ 2 )a
angular acceleration = 12.96 rad/s^2

I now understand nearly everything with the exception of that 1/2 on the Ia(angular acceleration) side. Why does that need to be present? Is it a constant for an angular inertia of a circle.


Regardless I still appreciate the help I'm getting. It gets me thinking more about the concepts and the methods behind solving these problems.

Zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz... No offense, but homeschooling is much eaiser, i'm in homeschool. : )

Finally figured out the Tension in the last problem. I'll edit that part out now.