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Theory of Machines Test 2

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Theory of Machines Test 2
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  • Question 1
    2 / -0.33
    In a four-bar mechanism, the lengths of driver crank, coupler and follower link are 150 mm, 250 mm and 300 mm respectively. The fixed link length is L0. Find the values of Lo so as to make it crank-crank mechanism.
    Solution

    Concept:

    According to Grashof’s Law:

    (1) S + L ≤ P + Q (Class I mechanism)

    There will be at least one complete revolution between the two links.

    • If shortest link is fixed: Double Crank Mechanism
    • If any of the adjacent link of the shortest link if fixed: Crank rocker mechanism
    • If the link opposite to shortest link is fixed: Double Rocker Mechanism

    (2) S + L > P + Q (Class II mechanism)

    • Only double rocker mechanism is possible.

    Calculation:

    For crank-crank mechanism, the condition to be satisfied are

    (i) Shortest link must be fixed i.e. Lo is shortest link.

    (ii) S + L ≤ P + Q

    Lo + 300 ≤ 250 + 150

    Lo ≤ 100 mm
  • Question 2
    2 / -0.33
    A flywheel is fitted to engine has a mass of 400kg and radius of gyration 0.2m. If initially engine is at rest and starting torque of engine is 400Nm. What is kinetic energy (kJ) of flywheel after 10 seconds?
    Solution

    Concept:

    T = Iα

    ω2 = ω1 + αt

    Calculation:

    I = mk2 = 400 × (0.2)2 = 16 kg.m2

    T = 400 N.m

    T = Iα

    α=40016

    α = 25 rad/s2  

    ∴ ω2 = ω1 + αt

    As ω1 = 0, t = 10

    ∴ ω2 = αt = 25 × 10

    ∴ ω2 = 250 rad/s

    (KE)t=10=12Iω22

    (KE)t=10=12(16)×(250)2

    ∴ (KE) (t=10) = 500 kJ

  • Question 3
    2 / -0.33
    If angle of obliquity of pair of gear wheels is 18˚. What is minimum number of teeth on pinion if arc of approach or recess not less than the pitch?
    Solution

    Concept:

    Maximum arc of approach of pinion,

    Maxarcofapproach=MaxPathofcontactCos

    Maxarcofapproach=rpsincos

     Max arc of approach = rp tanθ

    Where, rp = radius of pinion

    Given condition,

    Arcofapproach=πdpZp

    rptan2πrpZp

    Zp2πtan

    For ∅ = 18°

    Zp2πtan18

     Zp ≥ 19.32

    Zp = 20 

  • Question 4
    2 / -0.33
    Two 20° pressure angle involute gears in mesh have a module of 10 mm. The larger has 50 teeth and the pinion has 13 teeth. The centre distance between them is
    Solution

    Concept:

    Centredistance=D+d2

    D = Diameter of Gear, d = diameter of pinion

    Calculation:

    Given:

    ZG = 50 (Number of teeth on gear), ZP = 13 (Number of teeth on pinion), m = 10 mm

    We know

    m=DZ(1)

    Now,

    For gear

    D = mZG = 10 × 50 = 500 mm

    For pinion

    d= mZP = 10 × 13 = 130 mm

    Centredistance=D+d2=500+1302

    Centre distance = 315 mm

  • Question 5
    2 / -0.33
    The controlling force in a spring controlled governor is 1500 N when the radius of rotation of the balls is 200mm and 887.5 N when it is 130 mm the mass of each ball is 8 kg by changing the initial tension the governor behaves as isochronous governor. Then
    Solution

    Explanation:

    Given:

    F = ar + b

    1500 = 200a + b

    887.5 = 130a + b

    ⇒ a = 8.75 N/mm

    ⇒ b = - 250 N

    Now,

    For isochronous governor, b = 0

    F = mrω2 = ar

    ∴ a = mω2

    8.75 × 1000 = 8ω2

    ω = 33.07 rad/s

    ω=2πN60

    ∴ N = 315.8 rpm

  • Question 6
    2 / -0.33
    The moment of inertia an airplane air screw is 20 kg.m2 and the speed of rotation is 1000 rpm clockwise when viewed from the front. The speed of the flight is 200 km per hour. Find the gyroscopic acceleration (rad2/sec2) of the air screw on the airplane when it makes a left-handed turn on a path of 150 m radius 
    Solution

    Explanation:

    Given:

    Ip = 20 kg – m2

    ωs=2π×100060=104.67rad/sec

    ωp=Vr=200×10003600×150=0.37rad/sec

    Now,

    Gyroscopic acceleration = ωs × ωp

    ∴ Gyroscopic acceleration = 104.67 × 0.37

    Gyroscopic acceleration = 38.7279 rad2/sec2

  • Question 7
    2 / -0.33
    The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration 0.6 m. It rotates at 1800 rpm clockwise, when looking from the stern. Determine the gyroscopic couple (in kN-m) if the ship travels at 100 km/hr and steer to the left in a curve of 80 m radius.
    Solution

    Concept: 

    Gyroscopic couple C = Iωωp

    Calculation:

    Given:

    Radius of gyration (k) = 0.6 m, v = 100 km/hr, N = 1800 rpm, Radius (R) = 80 m, mass (m) = 8 tonnes 

    Now,

    ω=2πN60=2π×180060=188.5rad/s

    Now,

    Mass moment of inertia of rotor:

    I = mk2 = 8000 × (0.6)2 = 2880 kg-m2

    Angular velocity of precession:

    ωp=VR=100×1000360080=5001440=0.3472rad/s

    Gyroscopic couple (C) = Iωωp 

    C = 2800 × 188.5 × 0.3472

    C = 188500 Nm

    ∴ C = 188.5 kN-m

  • Question 8
    2 / -0.33
    A machine tool requires a torque of 100 + 20 sin θ N-m of torque, where θ is the crank angle. The power is supplied by an electric motor at uniform rate. A flywheel with a moment of inertia of 20 kg-m2 is attached. The maximum acceleration of the flywheel and the corresponding crank position is
    Solution

    Explanation:

    T = 100 + 20 sinθ

    Tmean=12π02π(100+20sinθ)dθ=100N.m

    Now,

    Change in Torque (ΔT) = Torque developed by engine – Torque required by machine

    Change in Torque (ΔT)  = 100 – (100 + 20 sinθ)

    Change in Torque (ΔT) = - 20 sinθ

    Now,

    To get (ΔT)max,

    ddθ(20sinθ)=0

    ⇒ - 20 cos θ = 0 ⇒ θ = 90 or 270

    (ΔT)max = -20 × sin 270

    (ΔT)max = 20 N.m

    Now,

    (ΔT)max = I αmax

    ⇒ 20 = 20 × αmax

    αmax = 1 rad/s2 at 270° 

  • Question 9
    2 / -0.33
    In a gear box, two meshing gears (Gear and pinion) of involute profile have a module of 5 mm with pressure angle ϕ =20° and a Gear ratio of 3. The number of teeth on Gear is 60 and take addendum as one module. At the end of engagement, if the pinion is rotating with a speed of 20 rad/s, the sliding velocity is
    Solution

    Concept:

    Sliding velocity is the velocity of one tooth relative to its mating tooth along the common tangent at the point of contact. It is given by

    VS = (ω1 + ω2) PC

    Where PC is the distance between pitch point and the point of contact.

    At the end of engagement, the distance between the pitch point and point of contact will be Path of recess.

    PC=ra2r2cos2ϕrsinϕ

    Calculation:

    Given:

    m = 5 mm, T = 60, G.R = 3, ωP = 20 rad/s, ϕ = 20°, addendum = W = 1m = 5 mm,

    From the given,

    R = m T/2 = 5 × 60 /2= 300/2=150 mm;

    G.R = T/t ⇒ t = 20

    r = m t /2 = 5 × 20/2 = 100/2 =50 mm;

    ra = r + W = 50 + 5 = 55 mm;

    ωG = ωP/G.R = 20/3 rad/s;

    PC=552502cos22050sin20=11.49mm

    Now,

    Sliding velocity = (20 + 6.67) × 11.49 = 306.7 mm/s

    Sliding velocity = 30.7 cm/s 

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