Theory of Machines Miscellaneous
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List-I (Gear types) List-II (Application) A. Worm gears 1. Parallel shafts B. Cross helical gears 2. Nonparallel, intersecting shafts C. Bevel gears 3. Nonparallel, non intersecting shafts D. Spur gears 4. Large speed ratios
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A → 4, B → 3, C → 2, D → 1
Correct Option: C
A → 4, B → 3, C → 2, D → 1
- For the configuration shown, the angular velocity of link AB is 10 rad/s counter clockwise. The magnitude of the relative sliding velocity (in ms–1) of slider B with respect to rigid link CD is
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Given: ωAB = 10 rad/sec (CCW)
AB = 250 mm, BC = 250 √3, AC = 500
Direction of velocity of AB is perpendicular to link AB and direction of velocity of AB is parallel to line CD.
Also, direction of relative velocity of slides B with respect to C is in line with link BC.
∴ VC = 0
But VB = rAB × WAB = 250 × 10 × 10– 3 = 2.5 m/sec
VBC = VB – VC = 2.5 m/secCorrect Option: D
Given: ωAB = 10 rad/sec (CCW)
AB = 250 mm, BC = 250 √3, AC = 500
Direction of velocity of AB is perpendicular to link AB and direction of velocity of AB is parallel to line CD.
Also, direction of relative velocity of slides B with respect to C is in line with link BC.
∴ VC = 0
But VB = rAB × WAB = 250 × 10 × 10– 3 = 2.5 m/sec
VBC = VB – VC = 2.5 m/sec
- A rigid link PQ is 2 m long and oriented at 20° to the horizontal as shown in the figure. The magnitude and direction of velocity VQ, and the direction of velocity VP are given. The magnitude of VP (in m/s) at this instant is
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By Instantaneous center method;say ω = angular velocity of rod
vq = rq . ω ...(i)
vp = rp . ω ...(ii) Dividing equation (i) with equation (ii) we getvq = rq ; vp = rp .vq vp rp rq Now by sine law, rp = rq sin(45° + 20°) sin(70°) rp = sin(65°) rq sin(70°) So , vp = sin(65°) vq = 0.96 sin(70°) Correct Option: D
By Instantaneous center method;say ω = angular velocity of rod
vq = rq . ω ...(i)
vp = rp . ω ...(ii) Dividing equation (i) with equation (ii) we getvq = rq ; vp = rp .vq vp rp rq Now by sine law, rp = rq sin(45° + 20°) sin(70°) rp = sin(65°) rq sin(70°) So , vp = sin(65°) vq = 0.96 sin(70°)
- In a cam-follower mechanism, the follower needs to rise through 20 mm during 60° of cam rotation, the first 30° with a constant acceleration and then with a deceleration of the same magnitude. The initial and final speeds of the follower are zero. The cam rotates at uniform speed of 300 rpm. The maximum speed of the follower is
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ω = 2πN = 10π 60
Time taken to move 30°(π / 180) × 30 = (π / 6) = 1 s ω 10π 60
During this period follower move 20 mm = .02 m and u = 0From s = ut + 1 at2 2 0.02 = 1 × a × 
1 
2 2 60
a = 144 m / sec2v = 0 + 144 × 5 = 2.4 m / sec 300 
Correct Option: D
ω = 2πN = 10π 60
Time taken to move 30°(π / 180) × 30 = (π / 6) = 1 s ω 10π 60
During this period follower move 20 mm = .02 m and u = 0From s = ut + 1 at2 2 0.02 = 1 × a × 1 2 2 60
a = 144 m / sec2v = 0 + 144 × 5 = 2.4 m / sec 300
- In the mechanism given below, if the angular velocity of the eccentric circular disc is 1 rad/s, the angular velocity (rad/s) of the follower link for the instant shown in the figure is
[Note: All dimensions are in mm.]
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∆PQR ∼ ∆SRO
∴ PQ = PO .....(1) SR SO
(PO)2 = (PQ)2 + (OQ)2
∴ PQ = √(PO)² - (OQ)²
= √50² - 25² = 43.3 mm
From (1),43.3 = 45 + 5 SR 5
SR = 4.33 mm
VQ = VR = SR × ω
= 4.33 × 1 = 4.33 m/sωPQ = VPQ = 4.33 = 0.1 rad / sec PQ 43.3
Correct Option: B
∆PQR ∼ ∆SRO
∴ PQ = PO .....(1) SR SO
(PO)2 = (PQ)2 + (OQ)2
∴ PQ = √(PO)² - (OQ)²
= √50² - 25² = 43.3 mm
From (1),43.3 = 45 + 5 SR 5
SR = 4.33 mm
VQ = VR = SR × ω
= 4.33 × 1 = 4.33 m/sωPQ = VPQ = 4.33 = 0.1 rad / sec PQ 43.3
