Question

5.The motion of the piston of an automobile engine is approximately SHM. a) If the stroke of an engine (twice the amplitude) is 0.100m and the engine runs at 3500rev/min., compute the acceleration of the piston at the endpoints of its stroke. b) If the piston has a mass of 0.450kg, what net force must be exerted on it at this point? c) What are the speed and kinetic energy of the piston at the midpoint of its stroke? d) What average power is required to accelerate the piston from rest to the speed found in c)?

Answer 1

Answer:

Part a)

$a = 6716.8 m/s^2$

Part b)

$F = 3022.56 N$

Part c)

$v = 18.325 m/s$

So kinetic energy is given as

$K = 75.55 J$

Part d)

$P = 1.76 \times 10^4 W$

Explanation:

Part a)

Amplitude of motion is given as

$A = 0.050 m$

angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{3500}{60})$

$\omega = 366.5 rad/s$

Now for acceleration of piston we have

$a = \omega^2 A$

$a = (366.5)^2(0.05)$

$a = 6716.8 m/s^2$

Part b)

Net force on the piston is given as

$F = ma$

$F = 0.450(6716.8)$

$F = 3022.56 N$

Part c)

Speed of the piston at the mean position is given as

$v = A\omega$

$v = (0.05)(366.5)$

$v = 18.325 m/s$

So kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.450)(18.325)^2$

$K = 75.55 J$

Part d)

Average power is given as ratio of work done and time

$P = \frac{W}{t}$

here we know that

$t = \frac{T}{4}$

$t = \frac{2\pi}{4\omega}$

$t = \frac{2\pi}{4(366.5)}$

$t = 4.28\times 10^{-3} s$

So power is given as

$P = \frac{75.55}{4.28 \times 10^{-3}}$

$P = 1.76 \times 10^4 W$