Mechanisms and Machines Questions and Answers

Answer: B

The moment of inertia is 0.096 kg.m².

The flywheel energy relation links the fluctuation of energy to the coefficient of fluctuation of speed \(C_{s}\) and the mean speed:

\[ \begin{aligned} \omega_{mean} &= \frac{220 + 280}{2} = 250\ \text{rad/s} \\ \Delta E &= I\,\omega_{mean}^{2}\,C_{s} \\ I &= \frac{300}{250^{2} \times 0.05} = \frac{300}{3125} = 0.096 \end{aligned} \]

The stated speeds are used only for the mean speed; the target coefficient 0.05 is the design value the flywheel must achieve. Using \(\omega_{max}\) or \(\omega_{min}\) in place of the mean speed gives about 0.077 or 0.124, neither of which is offered; 0.113, 0.071 and 0.053 do not follow from the data.

Answer: D

The transmission angle is maximum when the crank angle is 180°.

In a four-bar mechanism with crank length a and fixed link d, the distance between the crank pin and the fixed pivot of the output link is

\[ BD^2 = a^2 + d^2 - 2ad\cos\theta \]

The transmission angle μ is the angle opposite BD in the triangle formed with the coupler and the output link, so it grows as BD grows. BD is longest when cos θ = −1, that is at θ = 180°, which gives the maximum transmission angle. At θ = 0° BD is shortest and the transmission angle is minimum. At 90° and 270° it takes intermediate values.

Answer: D

A part isolated from a mechanism must be in equilibrium.

In static force analysis, each link is taken out as a free body with all the forces and couples that act on it from neighbouring links, and from outside. If the mechanism as a whole is in equilibrium, every isolated part of it must be in equilibrium too, so the sums of forces and moments on it are zero. In dynamic analysis the same holds once the inertia forces are added, by D'Alembert's principle.

That requirement is what lets the unknown joint forces be found link by link. 'May', 'may or may not' and 'cannot' would make the free-body method useless.