How Do You Calculate the Force to Slow a Spinning Disc to a Stop?

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Homework Statement


I need a quick sanity check here please. (Sorry, my alphas keep coming out as power signs, they're not!).

To find the force to slow a spinning disc with known moment of inertia (0.0004kgm2) and known angular velocity (120pi rad/s) to a stop in 6s.

Here, the deceleration (\alpha) is 120pi/6=20 rad/s2. The radius of the disc is 6cm=0.06m


Homework Equations



Torque = rf (r=radius, f=force)

Also, torque=I\alpha

And \alpha=w/t where w=angular velocity (120pi) and t= the time to slow=6s.

This gives \alpha to be 20pi rad/s2.


The Attempt at a Solution



Setting rf=I\alpha

0.06f=0.0004 x 20pi

From this, the force is given to be 0.42N.

Could someone just check if this is right. I think I've done it ok, but its been some time. It kind of looks right to me, but part of me is thinking I've missed something. It seems too easy. Any help greatly appreciated.
 
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Looks good to me. (Assuming the force is applied tangentially.)

To get your latex to align with the rest of a sentence, use the inline tag (itex) instead of the regular tag (tex): \alpha versus \alpha.
 
Opps. I forgot to add the lump of 5g of putty 4cm from the centre of the disc. This makes the total moment of inertia (0.004+0.005x0.042.

Therefore 0.06f=0.000408x20pi

so f=0.427N

But, thank you if you think that is right :wink:
 
How did the lump of clay get there? If the clay and disk start out rotating together at the given speed, then your answer is fine.
 
Yes, the clay was added as an earlier part of the question, hence I originally only had the moment of inertia of the disc (0.0004) and then changed it for the moment of inertia of both the disc and clay (0.004 + 0.005 x 0.0016 where the distance of the clay from the centre is 4cm, and its mass 5g. As it is a single particle on the disc, its own moment of inertia I worked using mr2, unlike 1/2 mr2 for the disc as the disc is a solid rotating object whilst the particle inscribes a hollow track).
 
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