A piston (with mass M = 2.0 kg) in a car engine is in vertical simple harmonic m

Question

A piston (with mass M = 2.0 kg) in a car engine is in vertical simple harmonic motion with amplitude A = 5.0 cm. If the engine is running at 3710 rev/min. What is the maximum velocity (m/s) of the piston? What is the maximum acceleration (m/s2) of the piston? Suppose a small 100 g piece of metal were to break loose from the surface of the piston when it (the piston) is at the lowest point (z = -5.0 cm and v = 0). At what position (m) the piece of metal lose contact with the piston? What is the velocity (m/s) of the metal piece when it loses contact with the piston?

Explanation / Answer

eqquation for simple harmonic motion

x = A sin(t - )

V = dx/dt = A * cos(t - )

Given = 3710/(2*pi*60) rad/s = 9.841 rad/s

(a)

Maximum velocity is

V = A = 5 * 10^-2 * 9.841 = 0.493 m/s

(b)

Acceleration = dv/dt = -A2 sin(t - )

Maximum acceleration = A2 = 5/100 * 9.841^2 = 4.84 m2/s

(c)

It looses contact when x = 0 m, since the bottom part starts to decelerate but the top part will not and move with same velocity.

(d)

K of metal needs to be found

= sqrt(k/m)

initially mass = 2 kg

= 9.841 rad/s

k = ^2 * m = 9.841^2 * 2 = 193.69

k per unit wt. of mass = k/2 = 193.69/2 = 96.85

when the metal breaks off,

it will have two springs it will acts as if two springs k1 and k2 are in series

K1 = 96.85 * ( 2 - 0.1) = 184.02

K2 = 96.85 * 0.1 = 9.685

Keff = 1/(1/k1 + 1/k2) = 1/(1/184.02 + 1/9.685) = 9.21

new = sqrt(Knew/m) = sqrt(9.21/2) = 2.15

Velocity max = Anew = 5 * 2.15/100 = 0.1075 m/s

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