# Homework Help: Tangential Force Required to Slow Cylinder to a rest?

1. Nov 6, 2012

### ObviousManiac

1. The problem statement, all variables and given/known data
a)-The massive grinder for wood-pulp in a newsprint paper factory, which can render about a half Tonne of wood into a thin soup in less than a minute, is essentially a stone cylinder of radius 0.900 m and mass about 6.00 Tonnes. An enormous motor spins this cylinder on its axis while hydraulic cylinders press the pocket of wood against its rim. A typical rotational speed for the motor is 62.0 rpm (rotations per minute). Suppose that while the paper plant is in full operation there is a power outage due to an ice-storm. How much energy (in J) is available to grind wood just from the rotational kinetic energy of the grinding wheel?

I HAVE ALREADY SOLVED PART a,
NEED HELP WITH B:

b)-If the grinding wheel comes to rest in 80.0 ms what is the tangential force (in N) that the wood presents to the grinding wheel if this force is constant during the slowing down?

2. Relevant equations

KE = 1/2Iw2
F = ∆p/∆t ...?
t = Fr
∆w/∆t = α

3. The attempt at a solution
honestly don't know where to start.

2. Nov 6, 2012

### tiny-tim

Hi ObviousManiac!
First find the angular acceleration, using the standard constant acceleration equations but with θ ω and α instead of s v and a.

Then find the torque that produces that angular acceleration, and so find the force at the rim.

3. Nov 7, 2012

### ObviousManiac

Silly question - what is my θ?

4. Nov 7, 2012

### tiny-tim

there's no θ in this particular question, only the angular equivalent of u v a and t

5. Nov 7, 2012

### ObviousManiac

to find α, could I not do this?:

wf = wi + αt

wf = 0, so

-wi = αt

(where wi = 62 rpm/60 sec * 2π rad = 6.49 rad/s)

so

- 6.49 = α(.08)

6. Nov 7, 2012

### tiny-tim

looks good!

and so the tangential force is … ?

7. Nov 7, 2012

### ObviousManiac

T = mr2a

= 6000(.92)(-81.125)

= -394267.5

rF = -394267.5

.9F = -394267.5

F = -394267.5/.9

F = - 438075 N

.... Is this correct?

8. Nov 7, 2012

### tiny-tim

No, that's the wrong I.

Also, it would be easier (and quicker, and less likely to lead to mistakes) if you wrote all your equations at the start …

in this case, Fr = τ = Iα, so F = Iα/r

9. Nov 7, 2012

### ObviousManiac

isn't I = mr2 ?

So would it be

F = Iα/r

F = mr2α/r = mrα

?

10. Nov 7, 2012

### tiny-tim

yes , except I for a solid cylinder isn't mr2

you need to learn the moment of inertia of every common geometrical shape

11. Nov 7, 2012

### ObviousManiac

OH

I didn't even think about that... wow I'm really dropping the ball tonight.

So:

I = .5mr^2

F = Ia/r

f = .5mr^2a/r = .5mra

= (6000)(.9)(-81.125)/2

= -219037.5 N

12. Nov 7, 2012