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In the below figure, PC is the connecting rod and OC is the crank making an angle $$\theta $$ with the line of stroke PO and rotates with uniform angular velocity at $$\omega $$ rad/s. The Klien's acceleration diagram for determining the acceleration of the piston P is shown by quadrilateral CQNO, the acceleration of the piston is _________ when the crank OC and connecting rod PC are at right angles to each other.<br> <img src="/images/question-image/mechanical-engineering/theory-of-machine/1528355411-43.JPG" title="Theory of Machine mcq question image" alt="Theory of Machine mcq question image">

Correct Answer: Zero
Theory of Machine mcq solution image

In the below figure, PC is the connecting rod and OC is the crank making an angle θ with the line of stroke PO and rotates with uniform angular velocity at $$\omega $$ rad/s. The Klien's acceleration diagram for determining the acceleration of the piston P is shown by quadrilateral CQNO. The acceleration of the piston P with respect to the crank-pin C is given by
Theory of Machine mcq question image

In the below figure, PC is the connecting rod and OC is the crank making an angle $$\theta $$ with the line of stroke PO and rotates with uniform angular velocity at $$\omega $$ rad/s. The Klien's acceleration diagram for determining the acceleration of the piston P is shown by quadrilateral CQNO, if N coincides with O, then
Theory of Machine mcq question image

Evaluate the following:
$$\frac{{\cos 2\theta \cdot \cos 3\theta - \cos 2\theta \cdot \cos 7\theta + \cos \theta \cdot \cos 10\theta }}{{\sin 4\theta \cdot \sin 3\theta - \sin 2\theta \cdot \sin 5\theta + \sin 4\theta \cdot \sin 7\theta }}$$

The value of $$\frac{{\sec \theta \left( {1 - \sin \theta } \right)\left( {\sin \theta + \cos \theta } \right)\left( {\sec \theta + \tan \theta } \right)}}{{\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)}}$$        is equal to:

$$\frac{{{{\left( {1 + \sec \theta \,{\text{cosec}}\,\theta } \right)}^2}{{\left( {\sec \theta - \tan \theta } \right)}^2}\left( {1 + \sin \theta } \right)}}{{{{\left( {\sin \theta + \sec \theta } \right)}^2} + {{\left( {\cos \theta + {\text{cosec}}\,\theta } \right)}^2}}},$$        0°

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find angular acceleration of connecting rod in rad/s2.

The expression $$\frac{{{{\left( {1 - \sin \theta + \cos \theta } \right)}^2}\left( {1 - \cos \theta } \right){{\sec }^3}\theta \,{\text{cose}}{{\text{c}}^2}\theta }}{{\left( {\sec \theta - \tan \theta } \right)\left( {\tan \theta + \cot \theta } \right)}},$$        0°

The velocity of piston in a reciprocating steam engine is given by (where $$\omega $$ = Angular velocity of crank, r = Radius of crank pin circle, $$\theta $$ = Angle turned by crank from inner dead center and n = Ratio of length of connecting rod to the radius of crank)

The expression $$\frac{{\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta } \right)\left( {\cot \theta + 1} \right)\cos \theta }}{{\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)\left( {1 + \tan \theta } \right){\text{cosec}}\,\theta }} - 1,$$       0°

In a slider crank mechanism, the crank rotates with a constant angular velocity of 300 rpm, Length of crank is 150mm, and the length of the connecting rod is 600mm. Determine acceleration of the midpoint of the connecting rod in m/s2. Crank angle = 45° from IDC.