Question

A train A runs from east to west and another train B of the same mass runs from west to east at the same speed with respect to earth along the equator. Normal force by the track on train A is N₁ and that on train B is N2:

• A. N₁ > N2
• B. N1 < N2
• C. N1= N2
• D. the information is insufficient to find the relation between N₁ and N2.

Answer 1

Answer: (A)

N₁ > N2

Answer 2

The Earth rotates from west to east with an angular velocity $\omega_e$. Let the mass of the train be $M$ and the radius of the Earth be $R$. Let the speed of the train with respect to the Earth's surface be $v$. The corresponding angular speed of the train relative to the Earth's axis (if it were stationary on the surface) is $\omega_t = \frac{v}{R}$.

The normal force $N$ exerted by the track on the train is the apparent weight of the train. This apparent weight is given by the gravitational force minus the centrifugal force due to the rotation. The general formula for the normal force at the equator is: $N = Mg - M\omega_{net}^2 R$ where $\omega_{net}$ is the net angular velocity of the train with respect to the Earth's center.

For Train A:

Train A runs from east to west. This direction is opposite to the Earth's rotation (west to east). Therefore, the net angular velocity of Train A with respect to the Earth's center is: $\omega_{net, A} = \omega_e - \omega_t$ The normal force for Train A is: $N_1 = Mg - M(\omega_e - \omega_t)^2 R$

For Train B:

Train B runs from west to east. This direction is the same as the Earth's rotation. Therefore, the net angular velocity of Train B with respect to the Earth's center is: $\omega_{net, B} = \omega_e + \omega_t$ The normal force for Train B is: $N_2 = Mg - M(\omega_e + \omega_t)^2 R$

Now, we compare $N_1$ and $N_2$. Since $\omega_e$ and $\omega_t$ are positive values, it is clear that: $(\omega_e + \omega_t) > (\omega_e - \omega_t)$ Squaring both sides (since both terms are positive, assuming $\omega_e > \omega_t$, which is true for typical train speeds): $(\omega_e + \omega_t)^2 > (\omega_e - \omega_t)^2$ Multiplying by $MR$: $M(\omega_e + \omega_t)^2 R > M(\omega_e - \omega_t)^2 R$ Let $C_1 = M(\omega_e - \omega_t)^2 R$ and $C_2 = M(\omega_e + \omega_t)^2 R$. So, $C_2 > C_1$.

Now substitute these back into the expressions for $N_1$ and $N_2$: $N_1 = Mg - C_1$ $N_2 = Mg - C_2$

Since $C_2 > C_1$, subtracting a larger value $C_2$ from $Mg$ will result in a smaller normal force $N_2$. Therefore, $N_1 > N_2$.