Theory of Machines Miscellaneous
- The input link O2P of a four bar linkage is rotated at 2 rad/s in counter clockwise direction as shown below. The angular velocity of the coupler PQ in rad/s, at an instant when ∠O4O2P = 180°, is
PQ = Q4Q = √2a
O2P = O2O4 = a
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When ∠O4O2 P = 180°Now, ω3 = I12I23 = a ω2 I12I23 2a ω3 = 1 2 2
ω3 = 1 rad/sCorrect Option: C
When ∠O4O2 P = 180°Now, ω3 = I12I23 = a ω2 I12I23 2a ω3 = 1 2 2
ω3 = 1 rad/s
- In a four-bar linkage, S denotes the shortest link length, L is the longest link length, P and Q are the lengths of other two links. At least one of the three moving links will rotate by 360° if
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According to Grashoff’s Criteria.
S + L ≤ P + QCorrect Option: A
According to Grashoff’s Criteria.
S + L ≤ P + Q
- For a mechanism shown below, the mechanical advantage for the given configuration is
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Mechanical advantage = Load (output force) Effort (input force)
For a four bar linkage in toggle position, effort = 0
∴ Mechanical Advantage = ∞Correct Option: A
Mechanical advantage = Load (output force) Effort (input force)
For a four bar linkage in toggle position, effort = 0
∴ Mechanical Advantage = ∞
- In a certain slider-crank mechanism, lengths of crank and connecting rod are equal. If the crank rotates with a uniform angular speed of 14 rad/s and the crank length is 300 mm, the maximum acceleration of the slider (in m/s2) is _________.
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amax = 2rω²
(where θ = 0) i.e at Inner dead centre
= 2 × 0.3 × 14²
amax = 117.6 m/s²Correct Option: C
amax = 2rω²
(where θ = 0) i.e at Inner dead centre
= 2 × 0.3 × 14²
amax = 117.6 m/s²
- A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in figure. The crank is rotating with a uniform angular speed of 10 rad/s, counter clockwise. For the given configuration, the speed (in m/s) of the slider is _______.
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L = 240 mm
r = 60 mmn = L = 4 r
ω = 10 rad/scosθ = sinθ n
θ = tan−1(4) = 75.96°∴ VS = ωr sinθ + sin2θ 2n ∴ VS = 10 × 0.06 sin75.96° + sin151.92° 2 × 4
= 0.6 × 1.0288
= 0.6 m/sCorrect Option: A
L = 240 mm
r = 60 mmn = L = 4 r
ω = 10 rad/scosθ = sinθ n
θ = tan−1(4) = 75.96°∴ VS = ωr sinθ + sin2θ 2n ∴ VS = 10 × 0.06 sin75.96° + sin151.92° 2 × 4
= 0.6 × 1.0288
= 0.6 m/s