Question

Consider the motion of a point (or particle) on the circumference of a rolling circle. As the circle rolls, it generates the cycloid \(\mathbf{r}(t)=b(\omega t-\sin \omega t) \mathbf{i}+b(1-\cos \omega t) \mathbf{j}\) where \(\omega\) is the constant angular velocity of the circle and \(b\) is the radius of the circle. Find the maximum speed of a point on the circumference of an automobile tire of radius 1 foot when the automobile is traveling at 55 miles per hour. Compare this speed with the speed of the automobile.

Short answer

The maximum velocity of a point on the circumference of the tyre is approximately 1.414 times the velocity of the car. Thus, the maximum speed of a point on the circumference significantly exceeds the speed of the automobile.

Step-by-step solution

Conversion of Speed

Firstly, convert 55 miles per hour to feet per second since the radius is given in feet. The conversion factor is \(1 \text{ mile/hour} = 1.47 \text{ ft/sec}\). So, auto speed \(v = 55 \times 1.47 \text{ ft/sec}\).

Calculation of Angular Velocity

Angular velocity \( \omega \) can be computed by \( \omega = v/r \), where \( r = 1 \text{ ft}\) is the radius of the tyre. Substituting \( v \) and \( r \) in the equation, angular velocity can be calculated.

Computing Maximum Speed

The maximum speed of a point on the circumference of the tyre can be found by differentiating the given vector \( r(t) = b(\omega t - sin \omega t)i + b(1 - cos \omega t)j \) and finding its magnitude. As \( omega \) has been deemed a constant, hence, the vector \( r(t) \) can be differentiated with respect to time \( t \). Note that since \( max(\dot{r}) = b * sqrt((\omega^2)^2 + \omega^2) = b * \omega * sqrt(2) \), this needs to be calculated.

Comparing Maximum Speed with Auto's Speed

Finally, the calculated maximum speed from Step 3 needs to be compared with the speed of the automobile from Step 1