Theory of Machines Miscellaneous
- Consider the triangular formed by the connecting rod and the crank of an IC engine as the two sides of the triangle. If the maximum area of this triangle occurs when the crank is 75°, the ratio of connecting rod length to crank radius to
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Δ = 1 (PQ) (PR) sin A 2
Area will be maximum when A = 90°
i.e. PQR is a right angled triangle.
∴ Ratio of connecting rod length to crank radius,l = tan 75° = 3.732; r
l = 3.732 r
Correct Option: C
Δ = 1 (PQ) (PR) sin A 2
Area will be maximum when A = 90°
i.e. PQR is a right angled triangle.
∴ Ratio of connecting rod length to crank radius,l = tan 75° = 3.732; r
l = 3.732 r
- A uniform thin disk of mass 1 kg and radius 0.1 m is kept on a surface as shown in the figure. The spring of stiffness k1 = 400 N/m is connected to the disk center A and another spring of stiffness k2 = 100 N/m is connected at point B just above point A on the circumference of the disk. Initially, both the springs are unstretched. A assume pure rolling of the disk. For small disturbance from the equilibrium, the natural frequency of vibration of the system is _____rad/ s (round off to one decimal place).
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Given, Disc mass m = 1 kg, R = 0.1 m
K1 = 400 N/m, K2 = 100 N/m
By assuming pure rolling of the disc, Mc = 0 at I.C. at point (C)
Icθ̇̇ + K2(2R)θ × 2R + K1Rθ × R = 0
Ic = Idisc + mR2 = mR2 + mR2 2 = 3 mR2 = 3 × 0.12 = 1.5 × 10-2 2 2 
⇒ ωn = √533.3rad / sec
ωn = 23.094 rad / sec
ωn = 23.1 rad / secCorrect Option: A
Given, Disc mass m = 1 kg, R = 0.1 m
K1 = 400 N/m, K2 = 100 N/m
By assuming pure rolling of the disc, Mc = 0 at I.C. at point (C)
Icθ̇̇ + K2(2R)θ × 2R + K1Rθ × R = 0Ic = Idisc + mR2 = mR2 + mR2 2 = 3 mR2 = 3 × 0.12 = 1.5 × 10-2 2 2
⇒ ωn = √533.3rad / sec
ωn = 23.094 rad / sec
ωn = 23.1 rad / sec
- A slender uniform rigid bar of mass m is hinged at O and supported by two spr i ngs, wit h stiffnesses 3k and k, and a damper with damping coefficient c, as shown in the figure. For the system to be critically damped, t he ratio c / √km should be
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I0 = m
L2+ m 
L 
2 12 4 = mL2 + mL2 = 7 mL2 12 16 48
Torque about hinge ‘O’= 7mL2 θ̇̇ + 3k. L2 θ + k. 9L2 θ + C. L2 θ̇ = 0 48 16 16 16 ⇒ 7mL2 θ̇̇ + 12kL2 θ + CL2 θ̇ = 0 48 16 16 
2ξωn = C × 48 = 3C 16 × 7 m 7m ⇒ 2√km = 3 × √7 × C 7 6 C = 4√7 √km Correct Option: D
I0 = m
L2+ m L 2 12 4 = mL2 + mL2 = 7 mL2 12 16 48
Torque about hinge ‘O’= 7mL2 θ̇̇ + 3k. L2 θ + k. 9L2 θ + C. L2 θ̇ = 0 48 16 16 16 ⇒ 7mL2 θ̇̇ + 12kL2 θ + CL2 θ̇ = 0 48 16 16 2ξωn = C × 48 = 3C 16 × 7 m 7m ⇒ 2√km = 3 × √7 × C 7 6 C = 4√7 √km
- Figure shows a quick return mechanism. The cranks OA rotates clockwise uniformly. OA = 2 cm, OO' = 4 cm. The ratio of time for forward motion to that for return motion is
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Ratio of time of forward motion to return motion
= 180 + 2α = 180 + 60 = 240 = 2 180 − 2α 180 − 60 120
(Given OD = 2 cm, OO ' = 4 cm, sin α = 2/4 = 0.5 ⇒ α = 30°)Correct Option: B
Ratio of time of forward motion to return motion
= 180 + 2α = 180 + 60 = 240 = 2 180 − 2α 180 − 60 120
(Given OD = 2 cm, OO ' = 4 cm, sin α = 2/4 = 0.5 ⇒ α = 30°)
- A rod of length 1 m is sliding in a corner as shown in figure. At an instant when the rod makes an angle of 60 degrees with the horizontal plane., the velocity of point A on the rod is 1m/ s. The angular velocity of the rod at this instant is
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VA = O'A × ω
1 = 1 cos 60° × ω
ω = 2 rad/sCorrect Option: A
VA = O'A × ω
1 = 1 cos 60° × ω
ω = 2 rad/s
