# Torque from moment of inertia

1. Jan 14, 2010

### physisis

I have an assignment that asks to find a torque from given angular velocity, mass and moment of inertia.

For a turntable disk, mass is known as 0.04452199 kilograms

The Angular velocity should be 106.81 (rads/sec)

These values are from solidwork file (anyone familiar with Solidworks?)

Center of mass: ( meters )
X = -0.01474448
Y = -0.00659061
Z = 0.04376940

Principal axes of inertia and principal moments of inertia: ( kilograms *
square meters )
Taken at the center of mass.
Ix = (-0.70029829, 0.71385033, 0.00002079) Px = 0.00001872
Iy = (-0.71385033, -0.70029829, -0.00002111) Py = 0.00001875
Iz = (0.00000000, -0.00002962, 1.00000000) Pz = 0.00003073

Moments of inertia: ( kilograms * square meters )
Taken at the center of mass and aligned with the output coordinate system.
Lxx = 0.00001874 Lxy = -0.00000001 Lxz = 0.00000000
Lyx = -0.00000001 Lyy = 0.00001874 Lyz = 0.00000000
Lzx = 0.00000000 Lzy = 0.00000000 Lzz = 0.00003073

Moments of inertia: ( kilograms * square meters )
Taken at the output coordinate system.
Ixx = 0.00010596 Ixy = 0.00000431 Ixz = -0.00002873
Iyx = 0.00000431 Iyy = 0.00011371 Iyz = -0.00001284
Izx = -0.00002873 Izy = -0.00001284 Izz = 0.00004234

As far as I know, equation for torque is angular acceleration X moment of inertia,
but in this case, I don't have Angular acceleration but velocity. and Radius is unknown.

Is it possible to get a value of Torque from these information? I am keep looking for way to approach this question but I am having a hard time with it.
Any help will be appreciated.

Last edited: Jan 14, 2010
2. Jan 14, 2010

### RoyalCat

Note that the turntable is spinning through an axis that is offset from its center of mass. That means that relative to its center of mass, the contact forces at the pivot produce a torque.

I think that the spin angular momentum vector goes through a precession because of this offset, but I'm not sure.

If I'm right in my approach, then the key to solving this question is: $$\vec \tau=\frac{d\vec L}{dt}$$ where in this case, $$\vec L$$ would be the spin angular momentum vector.