# Man on a railroad car rounding a curve

• LCSphysicist
In summary: the angular momentum equation will have a nonzero angular velocity and will not be symmetric sculpture.
LCSphysicist
Homework Statement
.
Relevant Equations
The friction equations
"A man of mass M stands on a railroad car which is rounding an
unbanked turn of radius R at speed v. His center of mass is height L
above the car, and his feet are distance d apart. The man is facing the
direction of motion. How much weight is on each of his feet?"

I came five equations, and five incognits:
Call the inside side with index i, outside with o

Ni + No = Mg
Ni*μ = mi*v²/R
No*μ = mo*v²/(R+d)

mi + mo = M
No*d/2 = No*μ*l + Ni*d/2 + Ni*μ*l

I will not develop, but you know that we can came to a answer to both N without μ.

The book came to this equations:

Ni + No = Mg
No*μ + Ni*μ = Mv²/R
No*d/2 = No*μ*l + Ni*d/2 + Ni*μ*l

The bold equations is where we desagree, if i could say, i'd say that book's solution is summarized, but justified if d is small in comparation, while my solution is more complete, but in some cases desnecessary. Example, F = Gmm/(r+d)² is the right, but F = m*g is a nice aproximation.

But, i don't know...

Let ##S## be the center of mass and ##\boldsymbol F_l, \boldsymbol F_r## are the reactions from the trolley to the left and right leg respectively:
$$M\boldsymbol a_S=M\boldsymbol g+\boldsymbol F_l+\boldsymbol F_r.$$

The angle momentum equation
$$0=\boldsymbol r_l\times \boldsymbol F_l+\boldsymbol r_r\times \boldsymbol F_r,$$
here
##\boldsymbol r_r## is a vector from the center of mass to the right sole,
##\boldsymbol r_l## -- respectively

Introduce a coordinate frame moving with the trolley and expand the equations

etotheipi
Yes I agree with @wrobel, for an extended body the net radial force is proportional to the centre of mass radial acceleration, and the COM is performing circular motion at speed ##v## and radius ##R + \frac{d}{2}##. For this NII equation, it doesn’t matter where the force is applied on the extended body.

I assume ##d << R## so we can essentially ignore the ##\frac{d}{2}##.

If you hypothetically knew the net force on e.g. his foot, you could write down an equation for the circular motion of the centre of mass of his foot and this would evidently be at a different radius.

Last edited by a moderator:
Substitute ##\boldsymbol F_l## from the first equation to the second one:
$$\boldsymbol F_l=M\boldsymbol a_S-M\boldsymbol g-\boldsymbol F_r;$$
$$(\boldsymbol r_r-\boldsymbol r_l)\times \boldsymbol F_r=M\boldsymbol r_l\times(\boldsymbol g-\boldsymbol a_S)$$
From the last equation ##\boldsymbol F_r## is determined up to additional horizontal vector directed radially. The vertical component of ##\boldsymbol F_r## is determined correctly

LCSphysicist said:
The book came to this equations:

Ni + No = Mg
No*μ + Ni*μ = Mv²/R
No*d/2 = No*μ*l + Ni*d/2 + Ni*μ*l

The bold equations is where we desagree, if i could say, i'd say that book's solution is summarized, but justified if d is small in comparation, while my solution is more complete, but in some cases desnecessary. Example, F = Gmm/(r+d)² is the right, but F = m*g is a nice aproximation.

But, i don't know...

I agree with the book's solution, except:

1) ##R \rightarrow R + \frac d 2##, as pointed out above.

2) The specific friction force on each foot is indeterminate. ##\mu N## is the maximum possible force of static friction, but it's only the sum of the friction forces that matters.

jbriggs444
I believe this problem still needs some extra explanation.

Actually the angular momentum equation for a rigid body (for the man here) is written as follows
$$J_S\boldsymbol{\dot \omega}+\boldsymbol\omega\times J_S\boldsymbol\omega=\boldsymbol r_l\times\boldsymbol F_l+\boldsymbol r_r\times\boldsymbol F_r.\qquad (*)$$
Here ##J_S## is man's inertia operator about his center of mass. By the conditions of the problem the angular velocity is constant: ##\boldsymbol{\dot \omega}=0##. This kills the first term in the left side of (*).

Further we implicitly assume that the man stands as a symmetric sculpture: the vertical axis passing through his center of mass is a principal axis for ##J_S## that is ##J_S\boldsymbol\omega=\lambda\boldsymbol\omega.## This assumption kills the second term in the left side of (*)

If the man takes down one hand or inclines aside or takes nonsymmetric pose somehow else then ##\boldsymbol\omega\times J_S\boldsymbol\omega\ne 0##.

Last edited:
etotheipi

## 1. What is the concept of "Man on a railroad car rounding a curve"?

The concept refers to the physical phenomenon that occurs when a man is standing on a railroad car that is traveling along a curved track. The man will experience a centrifugal force that pulls him towards the outside of the curve.

## 2. Why does the man experience a centrifugal force while rounding a curve on a railroad car?

This is due to the inertia of the man's body. As the train turns, the man's body wants to continue moving in a straight line, but the train's movement forces him to move in a curved path. This results in a centrifugal force acting on the man.

## 3. How does the speed of the train affect the centrifugal force experienced by the man?

The faster the train is moving, the greater the centrifugal force experienced by the man. This is because the higher speed of the train results in a larger change in direction, causing a stronger force to be exerted on the man.

## 4. Is the centrifugal force experienced by the man the same as the centripetal force acting on the train?

No, they are not the same. The centrifugal force is the force that the man experiences as he is pulled towards the outside of the curve, while the centripetal force is the force that keeps the train on its curved path. They are equal in magnitude but act in opposite directions.

## 5. Can the centrifugal force experienced by the man be eliminated?

Yes, the centrifugal force can be eliminated by either decreasing the speed of the train or increasing the radius of the curve. This will result in a smaller change in direction and therefore a weaker centrifugal force acting on the man.

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