Question

A locomotive is propelled by a turbine whose axle is parallel to the axes of wheels. The turbine's rotation direction coincides with that of the wheels. The moment of inertia of the turbine rotor relative to its own axis is equal to $I = 240kg \cdot {m^2}$. If the additional force exerted by the gyroscopic forces on the rails when the locomotive moves along a circle of radius R = 250 m with velocity $v = 50km/h$ is ${F_{add}} = \dfrac{x}{{10}}kN \cdot m$, find the value of $x$. The gauge is equal to $l = 1.5m$. The angular velocity of the turbine equals n = 1500 rpm.

Answer 1

Hint:- To turn the locomotive along a circle, apart from the centrifugal an additional force is needed. This force comprises the moment of gyroscopic forces and acts in equal and opposite directions on the 2 rails.

Formula used:-
Additional force:
${F_{add}} = \dfrac{{{M_g}}}{l}$ …… (1)
where,
$l$is the gauge length.
${M_g}$ is the Moment of gyroscopic forces.
Moment of gyroscopic forces:
${M_g} = \dfrac{{2\pi INv}}{R}$ …… (2)
where,
It is the moment of inertia.
N is the angular velocity in rps.
$v$ is the velocity.
R is the radius of the circle.

Step-by-step answer:-
Given:
1. The moment of inertia $I = 240kg \cdot {m^2}$
2. Radius of circle R = 250 m
3. Angular velocity of the turbine N = 1500 rpm.
4. Gauge length $l = 1.5m$
5. Velocity $v = 50km/h$
6. Additional forces${F_{add}} = \dfrac{x}{{10}}kN \cdot m$

To find: The value of $x$.

Step 1 of 3:
Convert the given quantities in S.I. units. Express velocity in m/s:
$v = 50 \times \dfrac{5}{{18}}m/s$
Express the angular velocity of turbine in rps:
$N = \dfrac{{1500}}{{60}}rps$

Step 2 of 3:
Use equation (2) to find the moment of gyroscopic forces:
${M_g} = \dfrac{{2 \times 3.14 \times 240 \times (\dfrac{{1500}}{{60}}) \times (\dfrac{{50 \times 5}}{{18}})}}{{250}}$

$${M_g} = \dfrac{{2 \times 3.14 \times 240 \times 1500 \times 50 \times 5}}{{250 \times 60 \times 18}} \\\ {M_g} = 2093.3N \cdot {m^2} \\\$$

Step 3 of 3:
Use eq (1) to find the value of ${F_{add}}$:
${F_{add}} = \dfrac{{2093.3}}{{1.5}} \\\ {F_{add}} = 1.4kN \cdot m \\\$
This can be written as:
${F_{add}} = \dfrac{{14}}{{10}}kN \cdot m$
Comparing with the expression for ${F_{add}}$ given we obtain $x$:
$x = 14$

Final Answer
The value of $x$ is 14.

Note:- In questions like these, apply the knowledge of gyroscopic forces and remember that additional force comprises gyroscopic forces. Comparing the given expression for additional forces will give the unknown quantity.