Theory of Machines Miscellaneous
- 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
-
View Hint View Answer Discuss in Forum
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
- 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).
-
View Hint View Answer Discuss in Forum
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 concentrated mass m is attached at the centre of a rod of Iength 2L as shown in the figure. The rod is kept in a horizontal equilibrium position by a spring of stiffness k. For very small amplitude of vibration, neglecting the weights of the rod and spring, the undamped natural frequency of the system is
-
View Hint View Answer Discuss in Forum
Equation of motion, is
m( Lθ̇ ).Lθ + kx × 2 L = 0
⇒ m( Lθ̇ ).Lθ + k(2L.θ) × 2 L = 0
⇒ mL2θ̇̇ + 4kL2 θ = 0
⇒ θ̇̇ + √4k / mθ = 0∴ ωn = √ 4k m Correct Option: D
Equation of motion, is
m( Lθ̇ ).Lθ + kx × 2 L = 0
⇒ m( Lθ̇ ).Lθ + k(2L.θ) × 2 L = 0
⇒ mL2θ̇̇ + 4kL2 θ = 0
⇒ θ̇̇ + √4k / mθ = 0∴ ωn = √ 4k m
- A 1.5 kW motor is running at 1440 rev/min. It is to be connected to a stirrer running at 36 rev/ min. The gearing arrangement suitable for this application is
-
View Hint View Answer Discuss in Forum
Power of motor = 1.5 kW
N = 1440 rev/min
Speed of stirrer = 36 rev/minS.R.D = 1440 = 40 36
∴ Worm gear is used.Correct Option: D
Power of motor = 1.5 kW
N = 1440 rev/min
Speed of stirrer = 36 rev/minS.R.D = 1440 = 40 36
∴ Worm gear is used.
Direction: A compacting machine shown in the figure below is used to create a desired thrust force by using a rack and pinion arrangement. The input gear is mounted on the motor shaft. The gears have involute teeth of 2 mm module.
- If the drive efficiency is 80%, the torque required on the input shaft to create 1000 N output thrust is
-
View Hint View Answer Discuss in Forum
Given: Module m = 2,
D = 2 T
∴ D = 80 × 2 = 160 mm
2F = 1000, or F = 500 N
Let T1 be the torque applied by motor.
T2 be the torque applied by gear.
∴ Power transmission = 80%
T1 ω1 = 2T2 × ω2 0.8 or T1 = 2 × F × (D / 2) × ω2 0.8 ω1 = 2 × 500 × 0.16 × 1 × 1 = 25 N-m 2 0.8 4
Correct Option: B
Given: Module m = 2,
D = 2 T
∴ D = 80 × 2 = 160 mm
2F = 1000, or F = 500 N
Let T1 be the torque applied by motor.
T2 be the torque applied by gear.
∴ Power transmission = 80%T1 ω1 = 2T2 × ω2 0.8 or T1 = 2 × F × (D / 2) × ω2 0.8 ω1 = 2 × 500 × 0.16 × 1 × 1 = 25 N-m 2 0.8 4
