Theory of Machines Miscellaneous
- A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are VP and VR, in the x and y directions, respectively. The magnitude of the angular velocity of the body is
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2VP = 2VR = ωbody √3 a
VP = √3VR (a = 1 unit)
ωbody = 2VRCorrect Option: B
2VP = 2VR = ωbody √3 a
VP = √3VR (a = 1 unit)
ωbody = 2VR
- Consider a slider crank mechanism with non zero masses and inertia. A constant torque τ is applied on the crank as shown in the figure. Which of the following plots best resembles variation of crank angle θ versus time
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Correct Option: D
- A 4-bar mechanism with all revolute pairs has link lengths lf = 20 mm, lin = 40 mm, lco = 50 mm and lout = 60 mm. The suffixes 'f, in', 'co' and 'out' denote the fixed link, the input link, the coupler and output link respectively. Which one of the following statements is true about the input and output links?
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S + L < P + Q
20 + 60 < 40 + 50
⇒ 80 < 90
If smaller link is fixed both input and Output link execute full circular motion.Correct Option: A
S + L < P + Q
20 + 60 < 40 + 50
⇒ 80 < 90
If smaller link is fixed both input and Output link execute full circular motion.
- For the four-bar linkage shown in the figure, the angular velocity of link AB is 1 rad/s. The length of link CD is 1.5 times the length of link AB. In the configuration shown, the angular velocity of link CD in rad/s is
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For the given configuration VAB = VCD
∴ ωAB AB = ωCD CD⇒ ωCD = ωAB AB CD 1 × 1 1.5 = 2 rad/s 3 Correct Option: D
For the given configuration VAB = VCD
∴ ωAB AB = ωCD CD⇒ ωCD = ωAB AB CD 1 × 1 1.5 = 2 rad/s 3
- A simple quick return mechanism is shown in the figure. The forward to return ratio of the quick return mechanism is 2 :1. If the radius of the crank O, P is 125 mm, then the distance d(in mm) between the crank centre to lever pivot centre point should be
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2 = 360 - α 1 α
or 2α = 360 – α
or α = 120°∴ α = 60° 2
i.e. ∠PO2 O1 =30°
In ∆PO2O1tan 30° = O1P d ⇒ d = 125 = 219.29 tan 30° and sin 30° = 125 x ⇒ x = 125 = 250 1/2 Correct Option: D
2 = 360 - α 1 α
or 2α = 360 – α
or α = 120°∴ α = 60° 2
i.e. ∠PO2 O1 =30°
In ∆PO2O1tan 30° = O1P d ⇒ d = 125 = 219.29 tan 30° and sin 30° = 125 x ⇒ x = 125 = 250 1/2