Machine Kinematics Questions and Answers – Closely-coiled Helical Spring and Compound Pendulum

«
»

This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Closely-coiled Helical Spring and Compound Pendulum”.

1. In a squared and ground helical spring the effective number of turns is increased by
a) 1
b) 2
c) 1.5
d) 0
View Answer

Answer: b
Explanation: In a compression spring the ends can be plain, plain and ground, squared, squared and ground. The number of inactive turns in each case are 1.5,1,1 and 2 respectively.
advertisement

2. Frequency of the fluctuating load on a helical compression spring should be
a) less than natural frequency of vibration
b) twenty times the natural frequency of vibration
c) slightly greater than the natural frequency
d) twenty times the natural frequency of vibration
View Answer

Answer: d
Explanation: If the frequency of application of load matches the natural frequency of vibration, very high stresses are induced in the spring material causing fatigue failure. It can be avoided by keeping the natural frequency at least 20 times greater.

3. Two concentric springs with stiffness equal to 100 N/mm and 80 N/mm respectively, when subjected to a load of 900 N will deflect by
a) 9 mm
b) 11.25 mm
c) 5 mm
d) 31.5 mm
View Answer

Answer: c
Explanation: In this case, total stiffness Kt is given by,
1/Kt = K1 + K2
= 100+ 80 = 180N/mm

Therefore, deflection = Force/stiffness
= 900N/180N/mm
= 5 mm.

4. Stiffness of the spring can be increased by
a) increasing the number of turns
b) increasing the free length
c) decreasing the number of turns
d) decreasing the spring wire diameter
View Answer

Answer: c
Explanation: Stiffness of the spring can be decreased by increasing the number of turns and increased by decreasing the number of turns.

5. The stress induced in an extra full length leaf in case of a prestresed leaf spring is
a) 1.5 times that in graduated leaves
b) equal to that in graduated leaves
c) dependent on the ratio of the number of extra full length and graduated leaves
d) none of the mentioned
View Answer

Answer: b
Explanation: The leaves which are cut from original triangular leaf, are termed as graduated leaves. In actual practice some extra leaves with the same length as that of top leaf are added to increase the stiffness of the spring.
advertisement

6. Initial gap between two turns of a close coiled helical tension spring should be
a) 0.5 mm
b) based on the maximum deflection
c) 1 mm
d) 0
View Answer

Answer: d
Explanation: There is no gap between two turns of a close coiled helical tension spring.

7. Due to addition of extra full length leaves the deflection of a semi-elliptic spring
a) increases
b) decreases
c) does not change
d) is doubled
View Answer

Answer: b
Explanation: In actual practice some extra leaves with the same length as that of top leaf are added to increase the stiffness of the spring.

8. A connecting rod of mass 5.5 kg is placed on a horizontal platform whose mass is 1.5 kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The wires are equally spaced round the circumference of a circle of 125 mm radius. When the c.g. of the connecting rod coincides with the axis of the circle, the platform makes 10 angular oscillations in 30 seconds. Determine the mass moment of inertia about an axis through its c.g.
a) 0.198 kg-m2
b) 1.198 kg-m2
c) 2.198 kg-m2
d) 3.198 kg-m2
View Answer

Answer: a
Explanation: Given : m1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m
Since the platform makes 10 angular oscillations in 30 s, therefore frequency of oscillation,
n = 10/30 = 1/3 Hz
Let kG = Radius of gyration about an axis through the c.g.
We know that frequency of oscillation (n)

1/3 = r/2πkG √g/l = 0.125/2πkG√9.81/1.25 = 0.056/kG

kG = 0.056 x 3 = 0.168 m
and mass moment of inertia about an axis through its c.g.,
I = mk2G = (m1 + m2)k2G
= (5.5 + 1.5) (0.168)2 kg-m2
= 0.198 kg-m2.

9. In order to find the radius of gyration of a car, it is suspended with its axis vertical from three parallel wires 2.5 metres long. The wires are attached to the rim at points spaced 120° apart and at equal distances 250 mm from the axis. It is found that the wheel makes 50 torsional oscillations of small amplitude about its axis in 170 seconds. Find the radius of gyration of the wheel.
a) 168 mm
b) 268 mm
c) 368 mm
d) 468 mm
View Answer

Answer: b
Explanation: Given : l = 2.5 m ; r = 250 mm = 0.25 m ;
Since the wheel makes 50 torsional oscillations in 170 seconds, therefore frequency of oscillation,
n = 50/170 = 5/17 Hz

advertisement

Let kG = Radius of gyration of the wheel
We know that frequency of oscillation (n),

5/17 = r/2πkG √g/l = 0.25/2πkG√9.81/2.5 = 0.079/kG

kG = 0.079 x 17/5 = 0.268 m = 268 mm.

10. A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the radius of gyration of the rod about a vertical axis through the centre of gravity.
a) 107 mm
b) 207 mm
c) 307 mm
d) 407 mm
View Answer

Answer: a
Explanation: Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,
n = 20/40 = 0.5 Hz

Let kG = Radius of gyration of the connecting rod.
We know that frequency of oscillation (n),

0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k
kG = 0.0535/0.5 = 0.107 m = 107 mm.

11. A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the mass moment of inertia of the rod about a vertical axis through the centre of gravity.
a) 0.014 kg-m2
b) 0.015 kg-m2
c) 0.016 kg-m2
d) 0.017 kg-m2
View Answer

Answer: d
Explanation: Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,
n = 20/40 = 0.5 Hz

Let kG = Radius of gyration of the connecting rod.
We know that frequency of oscillation (n),

0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k
kG = 0.0535/0.5 = 0.107 m = 107 mm
We know that mass moment of inertia,
I = mk2G = 1.5(0.107)2 = 0.017 kg-m2.

Sanfoundry Global Education & Learning Series – Machine Kinematics.
To practice all areas of Machine Kinematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn