# Kinetics of Rotational Motion

1. Jun 4, 2012

### freshbox

1. The problem statement, all variables and given/known data
The tension answer I got is different from the answer. Can someone help me take a look at my working? Really appreciate your time in doing this. Thanks alot.

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2. Jun 4, 2012

### freshbox

And if my answer is right... I want to ask the question states that "Angular acceleration compound pulley B is 5rad/s^2"

From this statement, why my radius is not the wheel radius (0.1) but I have to use the axle radius?

The whole wheel B is pulling mass A. shouldn't I use the wheel radius instead of the axle to find the acceleration?

More support to my statement: Wheel B And Parcel = Linear variables same. So all the more I should use Wheel radius to find acceleration.

Last edited: Jun 4, 2012
3. Jun 4, 2012

### Staff: Mentor

Yes, you should use the radius 0.1 because that's where the rope pulling A attaches.

4. Jun 4, 2012

### freshbox

Yea you are right, I sub in the values and got the exact answer.

Can I say for such questions to find acceleration, always use the radius that the rope is attaching to?

5. Jun 5, 2012

### Staff: Mentor

When you want the acceleration of the rope (which is the acceleration of the attached mass) then you want to use the radius where the rope attaches.

6. Jun 5, 2012

### freshbox

With regards to this attached question. They are asking me to "calculate the angular acceleration produced on the wheel"

Why can't i take the wheel radius since the question says "acceleration produced on the wheel" ?

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7. Jun 5, 2012

### Staff: Mentor

Not sure what you mean. The angular acceleration is the same at any radius. (Don't confuse angular with linear acceleration.)

There are two forces of interest acting on the wheel. The torque produced by each force depends on where it is applied--that's the radius you'd use to calculate the torque.

8. Jun 5, 2012

### freshbox

From Equation 3, how come I need to use the axle radius but not the wheel radius, since the question is asking me to find the angular acceleration of the wheel.

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9. Jun 5, 2012

### Staff: Mentor

Again, do not confuse angular acceleration of the assembly, which is the same for wheel and axle, with the linear (or tangential) acceleration of some part of the wheel, which depends on the radius at that point.

In Equation 3, you are relating the linear acceleration a from Equation 2 to the angular acceleration. The linear acceleration is the acceleration of the rope and hanging mass. Where is the rope attached?

10. Jun 5, 2012

### freshbox

rope is attached to the axle.

Last edited: Jun 5, 2012
11. Jun 5, 2012

### Staff: Mentor

Right. Thus to relate the linear acceleration of the rope (and thus the hanging mass) to the angular acceleration, you'll use the radius of the axle.

12. Jun 5, 2012

### freshbox

Does the wheel turn anticlockwise(because force 50N is pushing) and axle turning clockwise because Box A is pulling it down. Both turning different direction or the Wheel+Axle is turning together?

13. Jun 5, 2012

### Staff: Mentor

The wheel + axle is a single connected assembly. It turns as one unit. (Thus wheel and axle have the same angular acceleration.)

14. Jun 5, 2012

### freshbox

Ok i understand it already, thanks alot. Doc Al I have 1 last question, can you help me take a look.

I managed to find all the answers except for part E.

This is part of the working. My equation for FBD of Box C is as follow:

Summation Fy=ma
Tc-Wc=Mca

I have values for Tc, a, after subbing it in I have.
147.945-Wc=Mc(0.13)
I left 2 unknown which I am unable to continue.

I have look through the working from the FBD of Wheel and Box A but I cannot try to form an equation to find either Mc or Wc, can you guide me please. Thanks.

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15. Jun 5, 2012

### Staff: Mentor

You really only have one unknown. How are weight and mass related?

16. Jun 5, 2012

### freshbox

Weight = mass x 9.81

mass = Weight / 9.81

Last edited: Jun 5, 2012
17. Jun 5, 2012

### Staff: Mentor

Right! Weight = mass X g.

So now you can solve for the mass of C.

18. Jun 5, 2012

### freshbox

Ahh.... I got it already, thanks for the guidance and your time