JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II DYNAMICS OF MACHINERY Previous Question Paper (March - 2022 - R16)

B.Tech | II | II | R16 | March - 2022

Answer PaperPaper Code: 134AUMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

What is meant by ‘angle of heel’ in case of a moving motor cycle taking a turn? How is it determined?

7 marks

Answer

The angle of heel is the angle at which a motorcycle must incline from the vertical towards the center of the turn to maintain equilibrium. This inclination is necessary to counteract the centrifugal force acting on the motorcycle, preventing it from overturning.

It is determined by considering the balance of moments about the point of contact between the tires and the ground. For a motorcycle of mass m moving at velocity v along a curve of radius r, the centrifugal force F = mv2/r acts at the center of gravity (height h). The weight mg acts vertically downwards. By taking moments about the ground contact point, the angle of heel θ is given by:

tan(θ) = v2 / (g × r)
b

A flywheel having a mass of 20 kg and a radius of gyration of 300 mm is given a spin of 500 rpm about its axis which is horizontal. The flywheel is suspended at a point 250 mm from the plane of rotation of the flywheel. Find the rate of precession of the wheel.

8 marks

Answer

Given data:

  • Mass (m) = 20 kg
  • Radius of gyration (k) = 0.3 m
  • Moment of inertia (I) = m × k2 = 20 × (0.3)2 = 1.8 kg·m2
  • Spin speed (N) = 500 rpm
  • Angular velocity of spin (ω) = (2 × π × 500) / 60 = 52.36 rad/s
  • Distance from plane of rotation (x) = 0.25 m

The gyroscopic couple acting on the flywheel is C = I × ω × ωp, where ωp is the rate of precession. This couple is balanced by the gravitational torque caused by the weight of the flywheel acting at a distance x from the pivot point:

Torque = m × g × x = 20 × 9.81 × 0.25 = 49.05 N·m

Equating the gyroscopic couple to the gravitational torque:

1.8 × 52.36 × ωp = 49.05
ωp = 49.05 / 94.248 = 0.520 rad/s

The rate of precession is 0.520 rad/s.

2

Describe, in detail, the dynamic analysis of a slider crank mechanism to determine the velocity and acceleration of the piston, and the angular velocity and angular acceleration of the connecting rod.

15 marks

Answer

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3

The turning moment diagram for a petrol engine is drawn to vertical scale of 1 mm to 6 N-m, and a horizontal scale of 1 mm to 1°. The turning moment repeats itself after every half revolution of the engine. The areas above and below the mean torque line are 305, 710, 50, 350, 980, and 275 mm2. The rotating parts amount to a mass of 40 kg at a radius of gyration of 140 mm. Calculate the co-efficient of fluctuation of speed if the speed of engine is 1500 rpm.

15 marks

Answer

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415 marks
a

Define the terms ‘thrust along the connecting rod’ and ‘Turning moment on the crankshaft’. Derive suitable expressions for them.

7 marks

Answer

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b

Find the maximum and minimum speeds of a flywheel of mass 3250 kg and radius of gyration 1.8 m, when the fluctuation of energy is 112 kN-m. The mean speed of the engine is 240 rpm.

8 marks

Answer

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515 marks
a

Describe, with a neat sketch, the method of operation of rope brake dynamometer.

7 marks

Answer

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b

A conical pivot bearing 150 mm in diameter has a cone angle of 120°. If the shaft supports an axial load of 20 kN and the coefficient of friction is 0.03, find the power lost in friction when the shaft rotates at 200 rpm, assuming uniform pressure and uniform wear.

8 marks

Answer

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615 marks
a

Prove that the ratio of tensions in the tight and slack sides of a band and block brake is given by:

Tn / T0 = ((1 + μ tan θ) / (1 - μ tan θ))n

where θ = half of the angle subtended by each block on the center of the drum; Tn = tension on the tight side after n blocks; T0 = tension on the slack side.

8 marks

Answer

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b

A flat foot step bearing 300mm in diameter supports a load of 8kN. If the coefficient of friction is 0.1, and the speed of the shaft is 80 rpm, find the power lost in friction, assuming uniform wear.

7 marks

Answer

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7

A Hartnell governor having a central sleeve spring and two right-angled bell crank levers operates between 290 rpm and 310 rpm for a sleeve lift of 15 mm. The sleeve and ball arms are 80 mm and 120 mm respectively. The levers are pivoted at 120 mm from the governor axis and mass of the ball is 2.5 kg. The ball arms are parallel at lowest equilibrium speed. Determine the: (a) load on the spring at maximum and minimum speeds and (b) stiffness of the spring.

15 marks

Answer

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815 marks
a

Prove that the ratio of two successive amplitudes in case of under-damped system is constant.

7 marks

Answer

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b

A shaft, 1.5 m long and supported by flexible bearings at the ends, carries two wheels each of 50 kg mass. One wheel is situated at the center of the shaft and the other at a distance of 375 mm from the center towards left. The shaft is hollow of external diameter 75 mm and internal diameter 40 mm. The density of the shaft material is 7700 kg/m3, and its modulus of elasticity is 200 GN/m2. Find the lowest whirling speed of the shaft, taking into account the mass of the shaft.

8 marks

Answer

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About this paper

This page contains the solved question paper with answers of DYNAMICS OF MACHINERY from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech ME II II, March - 2022, R16.

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