When to apply $I_c \underline{\omega} = \underline{M_c}$?

Question

I was solving an exercise the other day, about a rolling cylinder on an inclined plane. Initially the cylinder slides, but then it begins to roll and the problem wanted to know the velocity of the Center of Mass at the instant in which the rolling motion begun. I instinctively applied $I_c \underline{\omega} = \underline{M_c}$ (all the moments expressed with respect to the Center of Mass) and solved the problem correctly according to the solution. However, after having given it some thought I realized that what I did was wrong: in fact, you can only apply such equation with respect to a point of instantaneous rotation (that is, a point instantaneously in quiet) or, in extreme cases, with respect to the Center of Mass when the istantaneous point of rotation exists and is at a fixed distance from the center and moving with it (which is not my case, since the point of istantaneous rotation moved from its initial position to the point of contact with the plane).

I would really like to know if I am right or wrong and why.

As always, any comment or answer is highly appreciated!

Answer 1

The expression $ \underline{M}_c = \mathrm{I}_c \,\underline{\omega}$ is never correct. I think you forgot the time derivative of rotational velocity here. Also, the are Coriolis torques that relate to the rotation of the rigid body

$$ \underline{M}_c = \mathrm{I}_c \,\underline{\dot{\omega}} + \underline{\omega} \times \mathrm{I}_c \underline{\omega} \tag{1}$$

This relates the net moment on the body $\underline{M}_c$ to the angular acceleration of the body $\underline{\dot{\omega}}$ and it similar to Newton's 2nd law $\underline{F} = m\, \underline{\dot{v}}_c$. Both together form the Newton-Euler equations of motion for the rigid body.

This is derived from the differentiation of

$$ \underline{M}_c = \tfrac{\rm d}{{\rm d}t} \underline{L}_c = \tfrac{\rm d}{{\rm d}t} \left( \mathrm{I}_c \underline{\omega} \right) = \mathrm{I}_c \tfrac{\rm d}{{\rm d}t} \left( \underline{\omega} \right) + \tfrac{\rm d}{{\rm d}t} \left( \mathrm{I}_c \right) \underline{\omega} $$

Care must be taken to sum the moments about the center of mass, evaluate the mass moment of inertia tensor about the center of mass, and express all quantities on the same basis vectors (same orientation).

The only simplification for (1) is when the rotation $\underline{\omega}$ coincides with a principal axis of rotation. Then the second term is dropped, as is the case for the 2D planar projection of (1).

When considering an arbitrary point b not at the center of mass, then rotational equations of motion become slightly more complex

$$ \underline{M}_{b} = {\rm I}_{b}\,\underline{\dot{\omega}}+\underline{\omega}\times{\rm I}_{b}\,\underline{\omega}+\underline{c}\times m\,\underline{\dot{v}}_{b} \tag{2}$$

where $\underline{c}$ is the position of the center of mass relative to b and $\mathrm{I}_b$ is the mass moment of inertia summed at b and $\underline{\dot{v}}_{b}$ is the acceleration of point b.

The above is an advantage if the acceleration of b is known, like in the case of a pin joint. In that case Newton's second law can be simplified because the acceleration of the center of mass is no longer needed

$$ \underline{F} =m\left(\underline{\dot{v}}_{b}-\underline{c}\times\underline{\dot{\omega}}+\underline{\omega}\times\left(\underline{\omega}\times\underline{c}\right)\right) \tag{3}$$

Answer 2

Cylinder rolling in the inclined plane.

starting with the free body diagram , you obtain two equations

translation $$m\,\ddot s=-F\tag 1$$ rotation $$I_{\text{CM}}\,\ddot\varphi=F\,r+\tau\tag 2$$

now , if the cylinder slide on the inclined plane , the contact force F is equal zero, if the cylinder rolled on the inclined plane ,from the rolling condition you obtain additional equation

$$s=r\,\varphi\quad\Rightarrow\quad \ddot s=r\,\ddot\varphi\tag 3$$

with those three equations you obtain the solution

$$\ddot\varphi=\frac{\tau}{I_{\text{CM}}+m\,r^2}$$ you can obtain this solution if you take the sum of the torques about the contact point A, (where $~m\,\ddot s=0~$, becuse the instantaneous rotation).

$$I_A\,\ddot\varphi=\tau$$

where

$$I_A=I_{\text{CM}}+m\,r^2$$