UPSC 2020 Maths Optional Paper 1 Q8c — Step-by-Step Solution

15 marks · Section B

Constrained motion · Dynamics & Statics · asked 7× in 13 yrs · Read the full method →

Question

A four-wheeled railway truck has a total mass $M$ , the mass and radius of gyration of each pair of wheels and axle are $m$ and $k$ respectively, and the radius of each wheel is $r$ . Prove that if the truck is propelled along a level track by a force $P$ , the acceleration is $\dfrac{P}{M+\frac{2mk^2}{r^2}}$ , and find the horizontal force exerted on each axle by the truck. The axle friction and wind resistance are to be neglected.

Technique

Rolling constraint $a=r\alpha$ ; rotational equation per wheelset gives friction $f=mk^2a/r^2$ ; whole-truck linear equation $P-2f=Ma$ yields $a$ ; isolate one wheelset for the axle (bearing) force $X=ma+f$ .

Solution

Setup

The truck has two pairs of wheels-and-axle (two wheelsets), each of mass $m$ , radius of gyration $k$ (so moment of inertia $I=mk^2$ about its own axis), and wheel radius $r$ . The total mass of the whole truck (body $+$ both wheelsets) is $M$ . The wheels roll without slipping, so if $a$ is the linear acceleration and $\alpha$ the angular acceleration of a wheelset,

a=r\alpha\quad\Rightarrow\quad \alpha=\frac{a}{r}.

Let $f$ be the (forward) friction force exerted by the rail on each wheelset at the contact point — this is what spins the wheels up.

Step 1 — Rotational equation for one wheelset

Taking moments about the axle’s centre (the only horizontal forces with a moment about the centre are the rail friction $f$ at radius $r$ ; the axle bearing force acts through the centre and has no moment):

f\,r=I\alpha=mk^2\cdot\frac{a}{r}\quad\Rightarrow\quad f=\frac{mk^2 a}{r^2}. \tag{1}

Step 2 — Linear equation for the whole truck

The external horizontal forces on the entire truck are the propelling force $P$ (forward) and the two rail-friction forces $f$ at the two wheelsets (these are backward reactions on the system). The internal axle forces cancel. Hence

P-2f=Ma. \tag{2}

Substitute $f$ from (1):

P-\frac{2mk^2 a}{r^2}=Ma \quad\Rightarrow\quad P=\Big(M+\frac{2mk^2}{r^2}\Big)a .

Therefore

\boxed{\,a=\dfrac{P}{M+\dfrac{2mk^2}{r^2}}\,}

as required. $\blacksquare$

Step 3 — Horizontal force on each axle

Consider one wheelset (mass $m$ ) in isolation. The horizontal forces on it are: the forward force $X$ transmitted by the truck body through the axle bearing, and the backward rail friction $f$ . Newton’s second law for the wheelset:

X-f=ma\quad\Rightarrow\quad X=ma+f=ma+\frac{mk^2 a}{r^2}=ma\Big(1+\frac{k^2}{r^2}\Big).

Insert the value of $a$ :

X=\frac{m\big(1+\tfrac{k^2}{r^2}\big)P}{M+\tfrac{2mk^2}{r^2}} =\frac{mP\,(r^2+k^2)}{Mr^2+2mk^2}.

Answer

\boxed{\,X=ma\Big(1+\frac{k^2}{r^2}\Big)=\frac{mP\,(r^2+k^2)}{Mr^2+2mk^2}\ \text{(forward, on each axle)}\,}