Does a body behave as a point mass even at rest?

[jbriggs444]

I am trying an experiment but I don't know how can I apply constant force each time.

Possibly it is time to actually learn the math. It is not like $\sum F_\text{ext} = m_\text{tot} a_\text{cm}$ is terribly difficult.

Then you can do a thought experiment. Consider a bar-bell type object. A thin and nearly massless rod connecting two massive balls that are small compared to the length of the rod.

Apply (on paper) a small force on the right-hand ball for a small time.
Consider the resulting motion of the assembly. How fast does the center of mass move? In what direction does the center of mass move?

 

[vanhees71]

One should also be careful, which external force you consider. All you discuss here is about the motion in the homogeneous gravitational field close to Earth.

To that end consider the body under consideration (it can be an elastic or rigid solid body or even a fluid) as a discrete set of point particles, held together by interaction forces, which can be described as "pair forces", i.e., on the particle with the label $j$ the total force is given by
$$\vec{F}_j=\sum_{k \neq j} \vec{F}_{jk} + m_j \vec{g},$$
where $\vec{g}$ is the constant gravitational acceleration on due to the Earth (an approximation valid for motions close to Earth). The equations of motion for the particles thus read
$$m_j \ddot{\vec{x}}_j = \sum_{k \neq j} \vec{F}_{jk} + m_j \vec{g}.$$
Now we sum this equation over $j$ and use Newton's Third Law ("lex tertia"), according to which $\vec{F}_{jk}=-\vec{F}_{kj}$, i.e., if the interaction force on particle $j$ due to the presence of particle $k$ is $\vec{F}_{jk}$ then the interaction force on particle $k$ due to the presence of particle $j$ is opposite. This means that by summing over $j$ all the interaction forces cancel and you get
$$\sum_j m_j \ddot{\vec{x}}_j=M \ddot{\vec{x}}_{\text{cm}} =M \vec{g},$$
where
$$M=\sum_j m_j, \quad \vec{x}_{\text{cm}}=\frac{1}{M} \sum_j m_j \vec{x}_j.$$
Our calculation shows that indeed the center of mass moves like a point particle in the gravitational field of the Earth,
$$\ddot{\vec{x}}_{\text{cm}}=\vec{g}=\text{const}.$$
For all other forces, and even the gravitational interaction with the Earth when considering the position dependence, is not as simple, because then you have (treating the Earth as a "point particle" sitting fixed at the origin of the coordinate frame),
$$m \ddot{\vec{x}}_j = \sum_{k \neq j} \vec{F}_{jk} -\gamma m_j m_{\text{Earth}} \frac{\vec{x}_j}{r_j^3} \quad \text{with} \quad r_j=|\vec{x}_j|.$$
Then summing over $j$ gives
$$M \ddot{\vec{x}}_{\text{cm}} = -\gamma m_{\text{Earth}} \sum_j \frac{m_j \vec{x}_j}{r_j^3}=\vec{F}_{\text{ext}},$$
and $\vec{F}_{\text{ext}}$ cannot be expressed in terms of $\vec{x}_{\text{cm}}$!

 

[mark2142]

If you look at the pencil tip where the force is applied, you can see that because the pencil both translates and rotates that the pencil tip moves farther than it would under either effect alone. So the work done and the kinetic energy gained will be greater than it would be under either effect alone. A force with the same magnitude can have greater effect because it acts over a greater distance.

We need more energy. We do not need more force.

But if we apply the force to the end of the pencil, the same force will be applied to a point on the pencil that moves for a greater distance. That means that the same magnitude of force can do a greater amount of work. It also means that we will have to put more effort into pushing the pencil on its end to maintain the same magnitude of force.

Intuitively, it is easier to push hard on a bowling ball than it is to push hard on a golf ball. The golf ball moves away fast when you push on it hard. It is difficult to keep up.

Both bowling ball and golf ball get the same momentum. But the golf ball gets 100 times more energy.

I think I got you. Its because its difficult to keep up at the end of a pencil(less mass) than at COM(more mass) when applying the force on its end that we are unable to provide it same momentum. We are unable to apply force for the required time. That's is why my glass cone on water experiment is failing. It is not covering the same distance when applying force off center.
Similarly the work done also becomes less since force is applied for lesser distance and so reduced KE for rotating object. Difficult to do. Yes?

 

[mark2142]

It takes time to connect the dots. Thank you guys @hutchphd @jbriggs444 and everyone else. That was some new and interesting information not given in a book.