Machine Dynamics Questions and Answers – Effect of Gyroscopic Couple on a Naval Ship during Steering and Pitching

This set of Machine Dynamics Questions and Answers for Freshers focuses on “Effect of Gyroscopic Couple on a Naval Ship during Steering and Pitching”.

1. It is required to lift water at the top of a high building at the rate of 50 litres/min. Assuming the losses due to friction equivalent to 5 m and those due to leakage equal to 5 m head of water, efficiency of pump 90%, decide the kilowatt capacity of motor required to drive the pump.
a) 0.102 kW
b) 0.202 kW
c) 0.302 kW
d) 0.402 kW
View Answer

Answer: c
Explanation: Work to be done = wQHg
w = 1 Kg/litre
Q = 50 litres per min
H = 20 + 5 + 5 = 30m
Work output/min = 1 x 50 x 30 x 9.81 Nm/min
Input power = 1 x 50 x 30 x 9.81/ 0.9 x 0.9 x1000 x 60
= 0.302 kW.

2. A power screw is rotated at constant angular speed of 1.5 revolutions/sec by applying a steady torque of 1.5Nm. How much work is done per revolution? What is the power required?
a) 141.36 W
b) 241.36 W
c) 341.36 W
d) 441.36 W
View Answer

Answer: a
Explanation: Work per revolution = Tϴ
= 15 x 2п
= 94.24 Nm per revolution
P = Work/sec
= 94.24 x 1.5 = 141.36 W.

3. Power is transmitted by an electric motor to a machine by using a belt drive. The tensions on the tight and slack side of the belt are 2200 N and 1000 N respectively and diameter of the pulley is 600 mm. If speed of the motor is 1500 r.p.m, find the power transmitted.
a) 46.548 kW
b) 56.548 kW
c) 66.548 kW
d) 76.548 kW
View Answer

Answer: b
Explanation: Power = Tω
= Frω = Fv
F = T1 – T2 = 2200 – 1000 = 1200N
v = пDN/60 = п x 600 x 1500/1000 x 60 = 47.12 m/s
P = 1200 x 47.12/1000 = 56.548 kW.

4. The flywheel of an engine has a mass of 200 kg and radius of gyration equal to 1 m. The average torque on the flywheel is 1200 Nm. Find the angular acceleration of flywheel and the angular speed after 10 seconds starting from rest.
a) 3 rad/s
b) 4 rad/s
c) 5 rad/s
d) 6 rad/s
View Answer

Answer: d
Explanation: I = mk2 = 2000 kgm2
T = Iα
α = T/I = 1200/2000 = 0.6 rad/s2
ω = ω0 + αt = 0 + 0.6 x 10 = 6 rad/s.

5. Calculate the moment of inertia and radius of gyration of a solid sphere of mass 10 kg and diameter 6.5m about its centroidal axis.
a) 2.055 m
b) 3.055 m
c) 4.055 m
d) 5.055 m
View Answer

Answer: a
Explanation: I = 2/5 mr2 = 49 kgm2
k = 0.6325 x 3.25 = 2.055 m.
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6. Calculate the work done per minute by a punch tool making 20 working strokes per min when a 30 mm diametre hole is punched in 5 mm thick plate with ultimate shear strength og 450 Mpa in each stroke.
a) 10.69 kNm
b) 20.69 kNm
c) 30.69 kNm
d) 40.69 kNm
View Answer

Answer: a
Explanation: Force required to punch one hole = area shared x ultimate shear strength = пdt x Ss
Ss = п x 30 x 5 x 450/1000 = 212.05 kN
Work done/min = Average force x Thickness of plate x No. of holes/min
= 212.05/2 x 5/1000 x 20 = 10.69 kNm.

7. In latitude 25.0 S, SA (spin axis) of a FG (free gyro) is in position S40E and horizontal. Find the tilt after 6 hours.
a) 61.16 up
b) 61.16 down
c) 51.15 up
d) none of the mentioned
View Answer

Answer: a
Explanation: PZ= 65, ZX = 90, tilt after 6 hours = 90 – ZY =?
In triangle PZX,
Cos PX = Cos Z Sin PZ
Cos PX = Cos 40 Sin 65
Cos PX = 0.694272
PX = PY = 46.031
Again, in triangle PZX,
Cos P = – Cot PZ Cot PX
Cos P = – Cot 65 Cot 46.031
P or angle ZPX = 116.732
Angle XPY = 6 hours = 90. Thus, angle ZPY = 26.732
In triangle PZY,
Cos ZY = Cos P Sin PZ Sin PY + Cos PZ Cos PY
Cos ZY = Cos 26.732 Sin 65 Sin 46.031 + Cos 65 Cos 46.031
ZY = 28.839 and tilt = 61.16 up.

8. The steering of a ship means
a) movement of a complete ship up and down in vertical plane about transverse axis
b) turning of a complete ship in a curve towards right r left, while it moves forward
c) rolling of a complete ship sideways
d) none of the mentioned
View Answer

Answer: b
Explanation: Steering means is to turn the vehicle to either left or right.

9. When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be
a) to move the ship towards starboard
b) to move the ship towards port side
c) to raise the bow and lower the stern
d) to raise the stern and lower the bow
View Answer

Answer: a
Explanation: When the rotor rotates in the anticlockwise direction, when viewed from the stern and the ship is steering to the left, then the effect of reactive gyroscopic couple will be to lower the bow and raise the stern. When the ship is steering to the right under similar conditions, then the effect of reactive gyroscopic couple will be to raise the bow and lower the stern.


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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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