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KDPhysics

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- Homework Statement
- A man of mass M stands on a railroad car that 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 a distance d apart. How much weight is on each of his feet?

- Relevant Equations
- Torque equation

I saw that the solution states that the torque about the center of mass is zero, since the man does not rotate about its center of mass.

However, I then thought about taking the torque about the left foot (so the right foot for the man's POV). Hence:

$$\tau_{left} = \tau_{0} + \textbf{R}\times \textbf{F}$$

where ##\textbf{R}## is the vector pointing from the left foot to the center of mass, ##\tau_{0}## is the torque about the center of mass (so zero) and ##\textbf{F}## is the net force on the man.Then:

$$N_R d - Mg\frac{d}{2}= 0 + \textbf{R}\times \textbf{F} \implies N_R=\frac{Mg}{2} + \frac{LMv^2}{Rd} $$

which is correct. However, I wrote that the torque about the left foot is ##\frac{d}{2}(2N_R-Mg)##, but clearly the man isn't rotating so shouldn't it be equal to zero? Then I would get that ##N_R = \frac{Mg}{2}##. So why isn't the torque about the left foot zero.

EDIT: thinking about it, perhaps this might have to do with the fact that the left foot is not an inertial frame of reference, and that therefore we have to take into consideration the centrifugal force acting on the man. Then. I find that:

$$\tau_{left}=N_R d - Mg\frac{d}{2} - L\frac{Mv^2}{R}$$

which when equated to zero gives the desired result. But then, is my previous method (where I wrote ##\tau_{left} = \frac{d}{2}(2N_R-Mg)## incorrect?

However, I then thought about taking the torque about the left foot (so the right foot for the man's POV). Hence:

$$\tau_{left} = \tau_{0} + \textbf{R}\times \textbf{F}$$

where ##\textbf{R}## is the vector pointing from the left foot to the center of mass, ##\tau_{0}## is the torque about the center of mass (so zero) and ##\textbf{F}## is the net force on the man.Then:

$$N_R d - Mg\frac{d}{2}= 0 + \textbf{R}\times \textbf{F} \implies N_R=\frac{Mg}{2} + \frac{LMv^2}{Rd} $$

which is correct. However, I wrote that the torque about the left foot is ##\frac{d}{2}(2N_R-Mg)##, but clearly the man isn't rotating so shouldn't it be equal to zero? Then I would get that ##N_R = \frac{Mg}{2}##. So why isn't the torque about the left foot zero.

EDIT: thinking about it, perhaps this might have to do with the fact that the left foot is not an inertial frame of reference, and that therefore we have to take into consideration the centrifugal force acting on the man. Then. I find that:

$$\tau_{left}=N_R d - Mg\frac{d}{2} - L\frac{Mv^2}{R}$$

which when equated to zero gives the desired result. But then, is my previous method (where I wrote ##\tau_{left} = \frac{d}{2}(2N_R-Mg)## incorrect?

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