Question

Consider a car engine running at constantspeed. That is, the crankshaft of the en-gine rotates at constant angular velocity whileeach piston moves back-and-forth in its cylin-der according to the rules of simple harmonicmotion.2030 rpm6.95 cmSuppose the two extremal positionsxmaxandxminof a piston are 6.95 cm from eachother.When the crankshaft of the engine rotatesat 2030 rpm (revolutions per minute), what isthe maximal speed|v|maxof the piston?Answer in units of m/s.

Answer 1

Answer:

The value is  $|v| = 7.39 \ m/s$

Explanation:

From the question we are told that

    The angular speed is  $w = 2030 \ rpm = \frac{2030 * 2 * \pi }{ 60} = 212.61 \ rad/s$

    The distance between the minimum and maximum external position is  $d = 6.95 \ cm = 0.0695 \ m$

Generally the amplitude of the crank shaft is mathematically represented as

         $A = \frac{d}{2}$      

=>     $A = \frac{0.0695}{2}$    

=>     $A = 0.03475 \ m$

Generally the maximum speed of the piston is mathematically represented as

        $|v| = A * w$  

=>    $|v| = 0.03475 * 212.61$

=>    $|v| = 7.39 \ m/s$