# Inclined plane moment of inertia

1. May 1, 2010

### pat666

1. The problem statement, all variables and given/known data

A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizontal. The coefficient of kinetic friction is 0.25. A string attached to the block is wrapped around a flywheel has mass 25.0 kg and moment of inertia 0.500 kgm2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200 m from that axis.
a) What is the acceleration of the block down the plane
b) What is the tension in the string
2. Relevant equations

3. The attempt at a solution
Ok ive been having a lot of trouble with things involving moments of inertia. ive found the friction that opposes the motion to be 9.81N. but how do i calculate the opposing force that the flywheel provides? any help is appreciated, as always..

2. May 1, 2010

### ehild

Could you please show a picture?

ehild

3. May 1, 2010

### ehild

Could you please show a picture?

ehild

4. May 1, 2010

### pat666

How do you show a picture in a reply?? ive attached it as a word doc.

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5. May 1, 2010

### Squeezebox

Use the relationship $$\alpha$$=a/r and solve the torque equation for a.

6. May 1, 2010

### pat666

you mean mg-T=ma??

7. May 1, 2010

### Squeezebox

First sum up all of the forces for the block.

$$\Sigma$$Fx=sin$$\theta$$*Fg-Ff-T=ma

Then get the pulley

$$\Sigma\tau$$=I*$$\alpha$$

8. May 1, 2010

### pat666

Ok ill do that and report back with my answer, thanks.

9. May 1, 2010

### pat666

Hey is the answer a=1.123m/s^2 and T=14N thanks

10. May 2, 2010

### ehild

Yes, it is!

ehild

11. May 3, 2010

### 69camaro

Hey Pat,

Could you describe what you've done to come up with those answers? I couldn't follow.

Cheers

12. May 3, 2010

### pat666

why do you want to follow, more cqu peoples???

Last edited: May 3, 2010
13. May 3, 2010

which part?