Question

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance between right and left wheels is 1.5 m. What are the rotating speeds of each driving wheel as fractions of the drive shaft speed?

Answer 1

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = $\frac{V}{R}$

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = $\frac{u}{(R - d)}$

                  u = $V \times \frac{(R - d)}{R}$

                    = $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

         n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

            = $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

             = $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =  

                   = $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

                   = $\frac{V}{r} \times 0.1651$