Theory of Machines Miscellaneous
- A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 rad/s. If the total fluctuation of speed is not to exceed ± 2% the mass moment of inertia of the flywheel in kg-m² is
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ΔE = 400Nm
ωmean = 20 rad/s
CS = ± 2%= 2 × 2 = 0.04 100
ΔE = Iω²meanCS
400 = I(20)² × 0.04
I = 25 kg-m²Correct Option: A
ΔE = 400Nm
ωmean = 20 rad/s
CS = ± 2%= 2 × 2 = 0.04 100
ΔE = Iω²meanCS
400 = I(20)² × 0.04
I = 25 kg-m²
- A gear train is made up of five spur gears as shown in the figure. Gear 2 is driver and gear 6 is driven member. N2, N3, N4, N5 and N7 represent number of teeth on gears 2,3,4,5 and 6 respectively. The gear(s) which act(s) as idler(s) is/are
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ω2 = ω2 ω3 ω5 = N3 N5 N6 = N3N6 ω6 ω3 ω5 ω6 N2 N4 N5 N3N4
only Gear(5) is IdleCorrect Option: C
ω2 = ω2 ω3 ω5 = N3 N5 N6 = N3N6 ω6 ω3 ω5 ω6 N2 N4 N5 N3N4
only Gear(5) is Idle
- For the epicyclic gear arrangement shown in the figure, ω2 = 100 rad/s clockwise (cw) and ωarm = 80 rad/s counter clockwise (ccw). The angular velocity ω5 (in rad/s) is
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Given: ωarm = – 80 rad/sec = Y (CCW)
ω2 = 100 rad/sec = X + Y (CW)
∴ X = 100 – Y = 100 + 8 = 108 rad/sec (CW)Also, ωs = - 1 X + Y = - 1 (180) + (-80) 3 3
= – 140 rad/sec (CCW)
∴ ωs = 140 rad/sec (CW)
Note: CCW → Counter Clock Wise
CW → Clock WiseCorrect Option: B
Given: ωarm = – 80 rad/sec = Y (CCW)
ω2 = 100 rad/sec = X + Y (CW)
∴ X = 100 – Y = 100 + 8 = 108 rad/sec (CW)Also, ωs = - 1 X + Y = - 1 (180) + (-80) 3 3
= – 140 rad/sec (CCW)
∴ ωs = 140 rad/sec (CW)
Note: CCW → Counter Clock Wise
CW → Clock Wise
- In the figure, link 2 rotates with constant angular velocity ω2. A slider link 3 moves outwards with a constant relative velocity VQ/P, where Q is a point on slider 3 and P is a point on link 2. The magnitude and direction of Coriolis component of acceleration is given by
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Direction is obtained by rotating velocity vector through 90º in the direction of rotation of the link.
Correct Option: A
Direction is obtained by rotating velocity vector through 90º in the direction of rotation of the link.
- The number of degrees of freedom of a five link plane mechanism with five revolute pairs as shown in the figure is
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DOF = 3(n – 1) – 2j – h n = 5, j = 5
DOF = 3(5 – 1) – 2 × 5 = 2Correct Option: C
DOF = 3(n – 1) – 2j – h n = 5, j = 5
DOF = 3(5 – 1) – 2 × 5 = 2