Topic: Theory of Machines
The piston of a steam engine moves with simple harmonic motion. The speed of rotation of crank is 120 rpm with a stroke of 2 m. What is the velocity of piston when it is 0.5 m from the centre?
- A. 4 * pi * $\sqrt{3}$
- B. pi * $\sqrt{3}$
- C. 2 * pi * $\sqrt{3}$
- D. 3 * pi * $\sqrt{3}$
Answer: C. 2 * pi * $\sqrt{3}$
Explanation: The crank radius (r) is half the stroke, so r = 2 / 2 = 1 m. The position from the start of the stroke is x = 1.5 m (which is 0.5 m from the center). Using the displacement equation x = r(1 - cos(theta)), we have 1.5 = 1(1 - cos(theta)), yielding cos(theta) = -0.5, thus theta = 120 degrees. The velocity of the piston is given by v = r * omega * sin(theta). With omega = (2 * pi * 120 / 60), v = 1 * (4 * pi) * sin(120 degrees) = 4 * pi * ($\sqrt{3}$ / 2) = 2 * pi * $\sqrt{3}$ m/s.