Physics, 02.11.2020 16:50 jby

If the stroke (twice the amplitude) of an engine is 0.100 m and the engine runs at 3500 rpm , compute the acceleration of the piston at the endpoint of its stroke. If the piston has a mass of 0.450 kg , what net force must be exerted on it at this point

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a) The magnitude of the acceleration of the piston at the endpoint of the stroke of the engine is 6716.809 meters per square second.

b) A net force of 3022.564 newtons must be exerted on the piston at this point.

Explanation:

a) After reading the statement carefully, we see the stroke moving on Simple Harmonic Motion, whose equation of position in time is:

$x(t) =A\cdot \sin (\omega\cdot t)+x_{o}$ (Eq. 1)

Where:

- Amplitude of the piston, measured in meters.

$\omega$ - Angular frequency, measured in radians per second.

- Time, measured in seconds.

$x_{o}$ - Initial position of the piston, measured in meters.

x(t) - Current position of the piston, measured in meters.

Acceleration function is found by differentiating the expression twice:

$\dot x(t) = \omega\cdot A\cdot \cos (\omega\cdot t)$ (Eq. 2)

$\ddot x(t) = -\omega^{2}\cdot A \cdot \sin (\omega\cdot t)$ (Eq. 3)

Where $\dot x(t)$ and $\ddot x(t)$ are the current velocity and acceleration of the piston, measured in meters per second and meters per square second, respectively.

The piston experiments maximum accelerations at endpoints, whose magnitude is determined by the following component:

$a_{max} = \omega^{2}\cdot A$ (Eq. 4)

The angular frequency of the system is:

$\omega = 3500\,\frac{rev}{min} \times \frac{2\pi\,rad}{1\,rev}\times \frac{1\,min}{60\,s}$

$\omega \approx 366.519\,\frac{rad}{s}$

If we know that $\omega \approx 366.519\,\frac{rad}{s}$ and $A = 0.050\,m$ , then the magnitude of the maximum acceleration is:

$a_{max} = \left(366.519\,\frac{rad}{s} \right)^{2}\cdot (0.050\,m)$

$a_{max} = 6716.809\,\frac{m}{s^{2}}$

The magnitude of the acceleration of the piston at the endpoint of the stroke of the engine is 6716.809 meters per square second.

b) By the Second Newton's Law, the net force exerted on the piston at this point (), measured in newtons, is:

$F = m\cdot a_{max}$ (Eq. 5)

Where is the mass of the piston, measured in kilograms.

If we get that $m = 0.450\,kg$ and $a_{max} = 6716.809\,\frac{m}{s^{2}}$ , then the net force that must be exerted on it at this point is:

$F =(0.450\,kg)\cdot \left(6716.809\,\frac{m}{s^{2}} \right)$

$F = 3022.564\,N$

A net force of 3022.564 newtons must be exerted on the piston at this point.

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