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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.
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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.
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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.
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
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
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
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°