Torque question

1. Feb 6, 2010

BoldKnight399

A 51.0 kg uniform solid cylinder has a radius of .392m. If the cylinder accelerates at 9.2*10^-2 rad/s^2 as it rotates about an axis through the center, how large is the torque acting on the cylinder? Answer in N*m.

so I plugged into the equation:
T=F*d
so
T=(5.1*(9.2^-2)*.392
T=1.839

that was wrong. If anyone has any ideas what I did wrong and feels like sharing their wisdom, that would be great.

2. Feb 6, 2010

tiny-tim

Hi BoldKnight399!

(have a tau: τ and an omega: ω and an alpha: α )
uhh?

But you don't know what F is.

Use τ = Iα.

3. Feb 6, 2010

BoldKnight399

But what is I? Is it:
(1/12)mL^2
and if so...what is my L? I only know the radius.

4. Feb 6, 2010

tiny-tim

You need to learn every moment of inertia on this list … http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken]

Last edited by a moderator: May 4, 2017
5. Feb 6, 2010

BoldKnight399

ok wow this is really helpful except that I don't have the height of the cylinder. What do i do then?

6. Feb 6, 2010

magwas

First of all, (5.1*(9.2^-2)*.392) is only 0.1839264
However (51*(9.2*10^-2)*.392) = 1.839264
The cylinder is solid. Rotational inertia is dependent on the shape of the object and its mass distribution.
You should either look it up ( see http://hyperphysics.phy-astr.gsu.edu/Hbase/mi.html#cmi ), or compute it.

7. Feb 6, 2010

magwas

You don't need the height. Moment of inertia is independent of it.

8. Feb 6, 2010

BoldKnight399

Wait...i don't? o never mind. This all makes sense now. Thank you!