Will an object undergoing translation move faster than if it underwent both translation and rotation?

Question

So I'm revisiting my physics, as I'm looking to solidify my foundations. I was going over the chapters concerning rotation of rigid bodies and I just thought of this problem: If an impulse is applied to a rigid object in such a way that it undergoes translation, will this object move at a higher speed than if the object was made to undergo both translation and rotation(about it's center of mass)under the same impulse? The impulse can be a max distance d from its center in the second case, and in line with the center of mass in the first case. Ignoring friction

Answer 1

What are the impulse equations of motion?

Consider an impulse vector $\boldsymbol{J}$ acting on a point located at $\boldsymbol{d}$ from the center of mass.

As a result of this impulse, the body with mass $m$ and mass moment of inertia $\mathcal{I}$ experience a step in velocity $\Delta \boldsymbol{v}$ of the center of mass, and a step in rotational velocity $\Delta \boldsymbol{\omega}$.

The equations of motion are:

$$ \begin{aligned} m \Delta \boldsymbol{v} & = \boldsymbol{J} \\ \mathcal{I} \Delta \boldsymbol{\omega} & = \boldsymbol{d} \times \boldsymbol{J} \end{aligned} $$

So change in motion is

$$ \begin{aligned} \Delta \boldsymbol{v} & = \tfrac{1}{m} \boldsymbol{J} \\ \Delta \boldsymbol{\omega} & = \mathcal{I}^{-1} \left( \boldsymbol{d} \times \boldsymbol{J} \right) \end{aligned} $$

As you can see the resulting translation is only a function of the impulse, and not its location.

Now consider the motion of the point of impulse application.

$$ \begin{aligned} \Delta \boldsymbol{v}_{\rm d} & = \Delta \boldsymbol{v} + \Delta \boldsymbol{\omega} \times \boldsymbol{d} \\ & = \tfrac{1}{m} \boldsymbol{J} + \left(\mathcal{I}^{-1} \left( \boldsymbol{d} \times \boldsymbol{J} \right) \right) \times \boldsymbol{d} \\ \end{aligned} $$

and project this into 2D such that $\boldsymbol{d}$ is a distance $d$ along the x-axis, and the impulse $\boldsymbol{J}$ has magnitude $J$ along the y-axis, then

$$ \Delta v_d = \left( \frac{1}{m} + \frac{d^2}{\mathcal{I}} \right) J $$

What results is the although the point of application of the impulse does not change the speed of the center of mass, it changes the speed of the impulse point. The further way the impulse acts, the larger $d$ is, the higher the speed $\Delta v_d$ is.

Answer 2

"move faster" seems slightly ambiguous here.

The translational speed of the center of mass changes by the same amount. If starting from rest, the speed of the center of mass will be the same in all situations with the same net impulse.

What will differ is the energy transferred and the rotational speed of the object.

Answer 3

The intuition is that in the case of an impulse at the COM the speed is greater. But it is because we forget that in a practical situation the net force is greater in this case. That means: the impulse is not the same.

One example is a yo-yo toy. If it is let to start to move from the rest by the force of gravity, the velocity after a fraction of a second can be thought as resulting from an impulse ($F_g \Delta t$). But the net force is smaller than the weight of the toy, because there is some tension in the string.

On the other hand, if the same toy is let to fall with a loose string, it will get a bigger speed after the same $\Delta t$, and without rotate, because the net force is now its full weight.