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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!

Matteo Menghini
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2 Answers2

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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

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

John Alexiou
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  • Please define c. – John Darby Jun 21 '22 at 21:37
  • @JohnDarby - Sorry, I had and deleted it. I added it back. – John Alexiou Jun 21 '22 at 23:31
  • These relationships are for the body frame (non-inertial frame) fixed in the moving body, not relationships in the fixed (inertial) space frame. – John Darby Jun 22 '22 at 13:36
  • @JohnDarby - why did you think so? They are not. In the body frame $\underline{\omega} = 0$ and similarly for other local quantities. Even though b might be riding on the body, from an inertial point of view it has $\underline{\dot{v}}_b \neq 0$ whereas if it was for an body riding frame the point would be fixed and $\underline{\dot{v}}_b = 0$ – John Alexiou Jun 22 '22 at 16:22
  • @JohnDarby - all i am stating are the Newton-Euler equations, that are well known and are the extension of $F = m a$ for rigid bodies. Always measured from an inertial frame and always along common basis vectors. – John Alexiou Jun 22 '22 at 16:25
  • You are in the inertial frame but you express the change in angular momentum with respect to the body frame. To quote Goldstein, Classical Mechanics, with respect to center of mass dL/dt (space frame) = M. To use the body axes: dL/dt (space frame) = dL/dt (body frame) + ω x L. This allows you to use $I$ as constant in the body frame. (I incorrectly implied earlier that you are fixed IN the non-inertial frame which as you pointed out is wrong). – John Darby Jun 22 '22 at 17:28
  • @JohnDarby - the question wasn't about how to estimate ${\rm I}c$. To do so you need to transform from the body frame to the inertial frame with $$ {\rm I}_c = \mathtt{R},\mathrm{I}{\rm body} , \mathtt{R}^\intercal$$ given the rotation matrix $\mathtt{R}$. Also deliberately I stayed away from the angular momentum derivative which is a rather involved subject when not expressed on the center of mass. – John Alexiou Jun 22 '22 at 21:07
  • I see. Do you have a good reference for general 3D motion? My physics tests do not have quite enough for me to really connect all the pieces. Thanks. – John Darby Jun 22 '22 at 21:56
  • Read the notes for Rigid Body Dynamics which approaches the subject from a simulation standpoint. – John Alexiou Jun 22 '22 at 23:43
  • @JohnDarby - When it comes to angular momentum see this answer and the link above. – John Alexiou Jun 22 '22 at 23:52
  • Thank you. Will review. – John Darby Jun 23 '22 at 00:02
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    I think I understand from your referenced earlier answer. Also, your answer to https://physics.stackexchange.com/questions/104513/derivation-of-eulers-equations-for-rigid-body-rotation helped. So Ic is in the inertial (space axes) frame, not in the non-inertial (body axis) frame, correct? – John Darby Jun 23 '22 at 03:59
  • @JohnDarby - yes. All quantities must have common basis vectors. – John Alexiou Jun 23 '22 at 13:13
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Cylinder rolling in the inclined plane.

enter image description here

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

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

Eli
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  • Only if the plane is a frictionless surface will the body slip with no rotation. With friction. the body rolls and slips. The greater the force of friction the greater the rolling, and if the force of friction is sufficient there is pure rolling and no slipping. – John Darby Jun 22 '22 at 13:41