Speed resulting from tangentially applying force to solid spheres with different mass distributions

Question

Given are two solid spheres of the same size and weight. They both have their center of mass at their geometric center.

One of them (A), however, has most of its mass distributed near the center (heavy center, light shell),

while the other (B) has most of its mass at the shell (light center, heavy shell).

Now I apply a tangential force ($F$) to them (they are at rest, floating in empty space). (When the ball starts to move, let's assume the force will stay tangential to the ball, i.e., the spot on the ball's surface, that the force will be applied to, changes with rotation. So the directions of the force will continue to be the same as it was at the start.)

Assuming I put the same amount of work into both ($F \cdot s$), I guess A will spin faster than B, because of the different moments of inertia.

Also, both will not only gain rotation but translation too. Will they both have the same (translational) speed, or does the difference in mass distribution make a difference here too?

Answer 1

For equal total energy, and equal force, the total time will be different, leading to different translational speed. This is because the energy fraction in translation vs rotation will be different. And that will affect how much energy per time is needed to provide equal force.

For equal force for equal time (not equal energy), the speed must be the same. Another way to see this, is to imagine the force as being applied by some particle at the edge somehow colliding tangentially. The particle feels an opposite and equal force, and thus develops a momentum. The spheres, regardless of their content, must have the same total opposite momentum, so that the sphere+particle momentum remains 0. If the force is equal for equal time (not equal work), then both spheres have the same total linear momentum and mass, hence the same translational speed.

Edit: Notes on energy

Note $Fr = I\dot{\omega}$ and $F = m\dot{v}$.

$2E = I\omega^{2} + mv^2$

$2\dot{E} = 2I\omega\dot{\omega} + 2mv\dot{v}$

$= 2\omega Fr + 2vF$

$= 2F(\omega r + v)$

What is $\omega r + v$? The instantaneous velocity of the part of the sphere on the very edge experiencing the force.

In time $\Delta{}t$ it moves a distance of $(\omega r + v)\Delta{}t$ so the work done is $\dot{E}\Delta{}t = F(\omega r + v)\Delta{}t = F\Delta{}x$.

In the particle analogy, note that in order for a stream of particles to provide the same force by collisions, the particles must have higher velocity commensurate with the edge velocity $\omega{}r+v$. It requires higher energy particles to provide the same force to a faster moving edge, so energy conservation is not broken--it is just that the energy input is higher to provide the same $F$, in a case where $\omega r + v$ is higher. But with Force and time specified, rather than total energy, it intuitively hold that the translational velocity of both spheres should be the same. With Power (energy/time) specified, the spheres would differ, as they would have the same energy at time $t$ but a different fraction of the energy would be rotational vs translational.

Edit 2:

I will try to argue that the relevant $F\cdot{}s$ is the edge of the ball, not the center of mass of the ball, and it depends on the velocity of the edge. Let us use a pole to fix the ball to rotate on its center of mass, so that $s_{COM}=0$. The COM clearly does not move. The pole fixing the COM to not move provides a force but moves no distance, and does no work. The force $F$ is spinning the ball and increasing the energy of the ball and doing work on the ball, but $F \cdot s_{COM}=0$. So if you believed that the relevant distance is the COM distance, you would find a contradiction. If you believe instead that $F\omega r dt$ is the instantaneous energy added because the edge moves with velocity $\omega r$ over a distance $\omega r dt$, then you can integrate. $E = \int F\omega r dt = \int F \dot{\omega}trdt = \int I\dot{\omega}^{2}tdt = \frac{1}{2}t^{2}I\dot{\omega}^{2} = \frac{1}{2}I\omega^{2}$

Assume $\dot{\omega}t = \omega$ (constant acceleration) and $Fr = I\dot{\omega}$ (torque definition)

Answer 2

The translation of the CG, which is at the geometric center for both spheres, is governed by ${\bf F} = m {\bf a}$, where ${\bf a}$ tracks the CG. You might say that since both spheres have the same mass, they will translate in the same way, but ${\bf F}$ becomes different for each sphere as time evolves.

If you attach a unit tangent vector ${\bf e}_t$ to the rim of the sphere, then ${\bf F}$ may be represented as ${\bf F} = F {\bf e}_t$. $F$ is the same for either sphere, but the evolution of ${\bf e}_t$ will depend on how quickly each sphere rotates. The sphere with the higher moment of inertia, for example, will rotate slower, and so ${\bf F}$ will spend more time initially vertical than in the case of the other sphere.

Therefore, the translation is different for each sphere.

Edit: Sorry I misunderstood the question. My answer is for the case where there is a follower force that is applied at the same material point for all time. The question was about a dead load applied to different material points in time.

Answer 3

The two spheres differ by their mass moment of inertia and have the same mass, and thus differ by their radius of gyration $g$.

Now consider where the center of rotation is for this situation. In the figure below the COR is located at point 0.

There is a special relationship between the distance to the COR from the center of mass $a$ and the distance where the force (or impulse) is applied $b$. This relationship is always

$$ \boxed{ a = \frac{g^2}{b} } \tag{1}$$

This means that if the center of mass has velocity $v_A$, then the point where the force is applied must have velocity $v_B = \frac{a+b}{a} v_A$ or dependent on the ratio of geometric radius $b$ to radius of gyration $g$

$$v_B = \left( 1 + \left( \tfrac{b}{g} \right)^2 \right) v_A \tag{2}$$

Analysis

Consider the force applied for some small time $\Delta t$ and the equations of motion are $$ \begin{aligned} m \frac{v_A}{\Delta t} & = F \\ (m g^2) \frac{\omega}{\Delta t} & = b F \end{aligned}$$ which yields the relationship between linear and rotational speed $$ \omega = \frac{b}{g^2} v_A \tag{3}$$

Finally consider the total kinetic energy $K = \tfrac{1}{2} m v_A^2 + \tfrac{1}{2} I \omega^2$ equals the work done $W=F s$ where $s$ is the distance traveled by force and is fixed for both scenarios.

$$ F s = \tfrac{m v_A^2}{2} + \tfrac{m b^2 v_A^2}{2 g^2} \tag{4} $$

Use (2) to solve for $v_A$ and substitute above. Then solve for $v_B$ to get the tangential velocity after the force has traveled the distance $s$ as a function of $m$ and $g$

$$ v_B^2 = \frac{2 F s}{m} \left( \frac{b^2+g^2}{g^2} \right) \tag{5} $$

Also note that the speed of the center of mass is

$$ v_A^2 = \frac{2 F s}{m} \left( \frac{g^2}{b^2+g^2} \right) \tag{6}$$

$$ \omega^2 = \frac{2 F s}{m} \left( \frac{b^2}{g^2 ( b^2+g^2)} \right) \tag{7}$$

Summary

As radius of gyration $g$ increases ↑ (more mass towards the outside) the following happens:

• The speed of the outer edge decreases ↓ (eq. 5)
• The speed of the center of mass increases ↑ (eq. 6)
• The spin decreases ↓ (eq. 7)

Answer 4

Conceptual Assessment of the Situation

First, in order to provide rotation (which it will) the force needs to move a bit with the surface. Otherwise the only energy transferred by the force will be along the distance of total translation, which cannot account for the work energy that created the angular kinetic energy.

At first blush, it seems this could be assumed to be from perfectly elastic collisions (elastic meaning no loss of energy during collision, and hence no sound energy or heat energy generated, and the net energy transfer from the change in velocity - magnitude that is - of the particles) from infinitely-many infinitesimal objects or a fluid jet. However, that doesnt work because it’s impossible to apply a tangential force that way.

One somewhat reasonable view of how this is happening is that a small charged bead is in the insulating ball just inside its surface, and the whole experiment is in a vertical electric field. The bead can move around the circumference freely. The problem is the dynamics of the bead transferring momentum to the ball. The real-world situation would be close.

That said, as with many such, the problem is theoretical (and technically theoretically impossible).

Concept of the Model

Mechanism aside, one important concept is that there can be no rotational energy imparted if the force only moves with the ball, ie as much as (and along a parallel path with) the movement of the center of mass of the ball. But we see due to the radial offset of the force from center that angular kinetic energy will develop.

Therefore, the conceptual model should be that a force F moves an infinitesimal distance $dy$ and then jumps back down that same $dy$ to do it again, staying ever tangential. Note that $dy$ comes from both the translation and the rotation so that the force is doing the two seemingly contradictory things we need it to do: 1. traveling with the surface so it can do all the needed work (enough to provide the energy for rotation and translation), and 2. remaining vertical. Calculations that only move the force the distance that the ball moves will necessarily violate conservation of energy.

Analysis

The torque is a set function of force, $Fr$, so the acceleration and angular acceleration are both proportional to force, the whole way. This constant ratio of $a$ and $\alpha$ means the angular and translational velocities will also be in that same proportion. Or, in math:

$$T=Fr=I \alpha,, F= \tfrac{I \alpha}{r}=ma$$ $$a = \frac{I}{mr} \alpha$$ $$v_0 , \omega_0 =0, a=k \alpha \implies v=k \omega, \forall t$$

(Because $v = \int adt=\int k \alpha dt = k \omega$ without integration constant. At the end of any duration $v_f=k \omega_f$ - meaning $\forall t$)

Kinetic energy:

$$E = \tfrac{1}{2} (mv^2+I \omega^2)= \tfrac{1}{2} [m( \frac{I}{mr} \omega)^2+I \omega^2]$$

$$= \tfrac{1}{2} (\frac{I^2}{mr^2} + I) \omega^2)$$

For the same $ \omega$’s, kinetic energy would be higher for case B, so $I_A <I_B,E_A=E_B \implies \omega_A > \omega_B$. This is not surprising. It says the ball with lower moment of inertia ends up spinning faster. Specifically $$ \frac{\omega_A}{\omega_B} = \sqrt{ \frac{ \tfrac{I_B^2}{m_Br_B^2} + I_B}{\tfrac{I_A^2}{m_Ar_A^2} + I_A}} = \sqrt{ \frac{I_B^2 + I_B m r^2}{I_A^2 + I_A m r^2}} $$ When $m_A=m_B,r_A=r_B, I_A<I_B$.

Subbing in $ \omega = \frac{mr}{I}v$ instead shows $v_A < v_B$. And we have a similar radical and ratio, except this time based on $m_i+ \tfrac{m_i^2r_i^2}{I_i}$.

Notes

There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$(which is how changes affect $v$), with $I_i+\tfrac{I_i^2}{m_ir_i^2}$(how changes affect $\omega$). For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass mean more energy is going into accelerating the object than before, with less available for rotation?

First, once under the radical, both include two terms rather than one for how it decreases with their respective inertial resistances ($m, I$). That’s because, for example, increasing $m$ (without changing $I$) has a double effect on $v$: one effect is because the portion of energy that goes toward translation now has to accelerate a larger mass (the $m^2$ in one term), and another effect is from the fact that larger $m$ makes less energy go toward translation as rotation has just become relatively easier (the $m$ term).

It is this latter factor, more energy going toward rotation whenever $m$ increases, that makes $\omega$ increase if $m$ goes up (the $m^{-1}$ term). Because $I$ hasn’t changed, the higher $\omega$ means a higher rotational kinetic energy, and because $E$ hasn’t changed, it also means a higher proportion of energy going to rotation. To summarize: Increasing $m$ increases the mass we need to accelerate linearly and decreases the energy going to translation, the double effect is the intuition for $m$ appearing in two terms in the $v$ ratio. The same for $I$ and $\omega$: $I$ appears in two terms in $\omega$ for the two effects, and is in one term for the single effect on $v$.

In some sense $m$ appearing twice in $v$‘s equation and once in $\omega$’s makes it possible for more overall motion to happen when $m$ is decreased, for the same $E$, (ie one will go up more than the other goes down). Similarly for $I$. This is not the case with radius. Remembering that $I$ is not changing, and increasing radius, we should not have a reduction in overall motion. Increases in $r$ make torque easier to create, but don’t increase energy or decrease overall resistance to motion. This means there will be more energy going to angular acceleration, but since $m$ and $I$ have not changed, $v$ drops a corresponding amount. Hence we have $r$ appearing only once in each ratio, as $r^2$.