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# Machine Kinematics MCQ – Double Hooke’s Joint

What is the purpose of double hooke’s joint?

a) Have constant linear velocity ratio of driver and driven shafts

b) Have constant acceleration ratio of driver and driven shafts

c) Have constant angular velocity ratio of driver and driven shafts

d) Have constant angular acceleration ratio of driver and driven shafts

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Answer: c

Explanation: The velocity of the driven shaft is not constant, but varies from maximum to minimum values. In order to have a constant velocity ratio of the driving and driven shafts, an intermediate shaft with a Hooke’s joint at each end is used.

Double hooke’s joint can be used to keep the angular velocity of the shaft constant.

a) True

b) False

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Answer: b

Explanation: Double hooke’s joint is used to keep the velocity ratio of driver shaft and driven shaft, It does not necessarily keeps the velocity constant.

Two shafts having an included angle of 150° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Using the above data, calculate the maximum angular acceleration of the driven shaft in rad/s2.

a) 6853

b) 6090

c) 6100

d) 6500

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Answer: a

Explanation: α = 180 -150 = 30⁰

cos2θ = 2sin2 α/1-sin2 α = 0.66

angular acc = dω/dt

= 6853.0 rad/s2.

Two shafts having an included angle of 150° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Using the above data, calculate the maximum torque required in N-m.

a) 822

b) 888

c) 890

d) 867

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Answer: a

Explanation: α = 180 -160 = 30⁰

cos2θ = 2sin2 α/1-sin2 α = 0.66

angular acc = dω/dt

= 6853 rad/s2

I = 0.12 Kg-m2

Therefore max torque = I.ang acc.

= 822 N-m.

Two shafts connected by a Hooke’s joint have an angle of 18 degrees between the axes.

Find the angle through which it should be turned to get the velocity ratio maximum.

a) 180

b) 30

c) 45

d) 90

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Answer: a

Explanation: Velocity ratio is ω1/ω = cosα/(1 – cos2θsin2α)

now this to be maximum cos2θ = 1

therefore θ = 0 or 180 degrees.

Two shafts connected by a Hooke’s joint have an angle of 18 degrees between the axes.

Find the angle through which it should be turned to get the velocity ratio equal to 1.

a) 30.6

b) 30.3

c) 44.3

d) 91.2

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Answer: c

Explanation: Velocity ratio is ω1/ω = cosα/(1 – cos2θsin2α)

now this to be 1

we get, cosα = 1 – cos2θsin2α

solving this equation we get

θ = 44.3 or 135.7 degrees.

Two shafts with an included angle of 160° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Find the maximum angular acceleration of the driven shaft.

a) 3090 rad/s2

b) 4090 rad/s2

c) 5090 rad/s2

d) 6090 rad/s2

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Answer: a

Explanation: Given : α = 180° – 160° = 20°; N = 1500 r.p.m.; m = 12 kg ; k = 100 mm = 0.1 m

We know that angular speed of the driving shaft,

ω = 2 π × 1500 / 60 = 157 rad/s

and mass moment of inertia of the driven shaft,

I = m.k2 = 12(0.1)2 = 0.12 kg – m2

Let dω1 / dt = Maximum angular acceleration of the driven shaft, and

θ = Angle through which the driving shaft turns.

We know that, for maximum angular acceleration of the driven shaft,

cos 2θ = 2sin2α/2 – sin2α = 2sin220°/2 – sin220° = 0.124

2θ = 82.9° or θ = 41.45°

and dω1 / dt = ω2cosα sin2θsin2α/(1 – cos2θsin2α)2

= 3090 rad/s2.

The angle between the axes of two shafts connected by Hooke’s joint is 18°. Determine the angle turned through by the driving shaft when the velocity ratio is maximum.

a) 90°

b) 180°

c) 270°

d) 360°

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Answer: b

Explanation: Given : α = 98°

Let θ = Angle turned through by the driving shaft.

We know that velocity ratio,

ω1/ω = cosα/1 – cos2θsin2α

The velocity ratio will be maximum when cos2 θ is minimum, i.e. when

cos2 θ = 1 or when θ = 0° or 180°.