What is the exact constraint for two bodies remain in contact? Consider the case of a rod constrained to move downward on an inclined wedge. It is known that the velocity of the wedge and the velocity of the rod in the common normal direction should be the same because they must not lose contact.

Differentiating this we can see that acceleration along common normal direction must be the same too.

Let us consider a curved surface. It is correct to say that the velocity of wedge and the velocity of the rod in the normal direction must be the same at any instant, but the same is no longer necessary for accelerations. Why is it so?

example problem- curved surface-rod constraint

Namely $v_w\sin \theta=v_r\cos \theta$, but differentiating this does not give $a_w\sin \theta=a_r\cos \theta$, here $\theta$ is the instantaneous angle made by the common normal with the horizontal. So how does one arrive to the contact constraint? And why is it the way it is?

P.S. It seems as though the rate of change of velocity in common normal direction isn't just the acceleration component in that direction but something more.


2 Answers 2


I had the same question exactly when I was learning dynamics (I think).

The question at hand is that if a constrained is formed in terms of velocities, how do we transform the constraint in terms of accelerations so that it can be used in ODE system, of equations.

The simplest example is that of a slipping wheel on a plane.


  • Velocity Kinematics

    It is correct to transfer the velocity to the contact point and impose the constraint there

    $$ \boldsymbol{\hat{j}} \cdot \boldsymbol{v}_A = 0 \tag{1} $$ where the velocity vector at A is evaluated from the velocity vector at C and the location vectors $\boldsymbol{r}_A$ and $\boldsymbol{r}_C$ $$\boldsymbol{v}_A = \boldsymbol{v}_C + \boldsymbol{\omega} \times ( \boldsymbol{r}_A - \boldsymbol{r}_C) \tag{2}$$

  • Material Acceleration (incorrect)

    It is incorrect to take a time derivative of (1) to establish the constraint in terms of accelerations.

    $$\require{cancel} \cancel{ \boldsymbol{\hat{j}} \cdot \boldsymbol{a}_A = 0 } \tag{3}$$ with $$ \boldsymbol{a}_A = \boldsymbol{a}_C +\underbrace{ \boldsymbol{\alpha} \times ( \boldsymbol{r}_A-\boldsymbol{r}_C)}_\text{Euler term} + \underbrace{ \boldsymbol{\omega} \times (\boldsymbol{v}_A - \boldsymbol{v}_C)}_\text{Coriolis term} \tag{4}$$

    The problem here is the Coriolis term which points away from the contact. Point A accelerates towards the center C and it is incorrect to force this term to zero. This is because $\boldsymbol{a}_A$ is the material acceleration of A, which means it tracks the acceleration of a point riding on the rigid body.

  • Spatial Acceleration (correct)

    Instead, look at the velocity of whatever point happens to be in contact, and how that changes with time. This is the spatial acceleration (a term familiar to fuild mechanics people). So take the partial derivative of (2), with only $\boldsymbol{v}_C$ and $\boldsymbol{\omega}$ changing (the motion of the body), but keep the location of A fixed. $$ \boldsymbol{a}_A^\star = \boldsymbol{a}_C + \underbrace{\boldsymbol{\alpha} \times ( \boldsymbol{r}_A - \boldsymbol{r}_C)}_\text{Euler term} \tag{5}$$ and impose the constraint $$ \boldsymbol{\hat{j}} \cdot \boldsymbol{a}_{A}^\star = 0 \;\; \checkmark \tag{6} $$


Velocity along the common normal direction must be the same as the distance between the bodies must remain constant, but the same may not be true for acceleration.

Consider a rod rotating with constant angular velocity. Different points on the rod have different accelerations along the normal direction but the velocities in that direction are same for all the points (zero). Simply because different acceleration along the normal direction (or more generally, along the line joining the two bodies) does not mean that the bodies will approach each other. Rather, it could mean that the direction of relative tangential velocity (tangential meaning, along the direction perpendicular to the line joining the two) is changing with time.

  • $\begingroup$ Yeah what you are saying is true, but why is it that rate of change of velocity in the common normal direction isn't simply the acceleration component in that direction? $\endgroup$ Aug 4, 2020 at 5:19
  • $\begingroup$ Although the acceleration constraint may be incorrect, it should still be right to say that component of acceleration in a given direction is the rate of change of velocity in that direction right? $\endgroup$ Aug 4, 2020 at 5:24
  • $\begingroup$ @RameshGupta It is, it's just that the common normal direction is also changing with time. $\endgroup$ Aug 4, 2020 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.