# Theory of Machines Questions and Answers – Centrifugal Clutch

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This set of Theory of Machines Multiple Choice Questions & Answers (MCQs) focuses on “Centrifugal Clutch”.

1. Which of the following clutches include shoes and spider inside the rim of the pulley?
a) Centrifugal clutch
b) Cone clutch
c) Multi plate clutch
d) Single plate clutch

Explanation: Centrifugal clutches are incorporated into motor pulleys which consist of a number of shoes on inside of the rim of the pulley. These clutches are generally incorporated into the motor pulleys. The outer surfaces of the shoe are covered with a layer of a friction material.

2. In a centrifugal clutch, what is ω?
a) Angular acceleration of the pulley
b) Angular running speed of the pulley
c) Angular acceleration at which the engagement begins to take place
d) Angular running speed at which the engagement takes place

Explanation: In a centrifugal clutch, ω is the angular running speed of the body. ω1 is the angular speed at which the engagement begins to take place.

3. What is the formula for the total frictional torque transmitted?
a) µ(Pc – Ps)R × n
b) µ(Pc – Ps)R
c)µ(Pc + Ps)R
d)µ(Pc + Ps)R x n

Explanation: Frictional torque acting on each shoe = µ(Pc – Ps)R. Therefore, total frictional torque transmitted = µ(Pc – Ps)R × n; where n is the number of shoes.
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4. In a centrifugal clutch, total frictional torque transmitted = µ(Pc – Ps)R × n; where Pc is the ____________________ acting on each shoe and is given by the formula Pc = _______
a) centrifugal force, mω12r
b) inward force, mω2r
c) centrifugal force, mω2r
d) inward force, mω12r

Explanation: Pc is the centrifugal force acting on each shoe at the running speed; Pc = mω2r.
A little consideration will show that Pc is greater than Ps. The net outward radial force of the shoe is given as Pc – Ps.

5. In a centrifugal clutch, total frictional torque transmitted = µ(Pc – Ps)R × n; where Ps is the ____________________ acting on each shoe and is given by the formula Ps = _______
a) centrifugal force, mω12r
b) inward force, mω2r
c) centrifugal force, mω2r
d) inward force, mω12r

Explanation: Ps is the inward force acting on each shoe at the running speed; Ps = mω12r. A little consideration will show that Pc is greater than Ps. The net outward radial force of the shoe is given as Pc – Ps.

6. If the radial clearance (c) is specified and is not negligible, then what is the operating radius of the mass centre of the show from the axis of the clutch?
a) r1 = r – c
b) r1 = c – r
c) r1 = r + c
d) r1 = r x c

Explanation: If the radial clearance (c) is specified and is not negligible, then the operating radius of the mass centre of the show from the axis of the clutch is r1 = r + c. So now, Pc = mω2r1.

7. In a centrifugal clutch, if l and b are the contact length and the width of the shoes respectively and p is the intensity of pressure exerted in the shoe, l.b.p = Pc– Ps. True or false?
a) True
b) False

Explanation: l.b.p = Area x p; which is also equal to the net outward force with which the show presses against the rim at running speed. Thus, l.b.p = Pc – Ps.

8. If the inside radius of the pulley rim is equal to 150 mm and the angle subtended by shoes at the centre of the spider is 60°, find the net outward radial force of the centrifugal clutch. The intensity of pressure is equal to 0.1 N/mm2 and the width is given to be 100 mm.
a) 1570.8 N
b) 1823.7 N
c) 1289.4 N
d) 1438.9 N

Explanation: θ = 60° = π/3 rad, R = 150 mm, b = 100 mm and p = 0.1 N/mm2
l = θ R = 157.08 mm
Net outward radial force of the centrifugal clutch = l.b.p = Pc – Ps
Pc – Ps = 157.08 x 100 x 0.1 = 1570.8 N.

9. A centrifugal clutch is to transmit 25 kW at 1200 r.p.m. There are four shoes. The speed at which the engagement begins is 0.5 times the running speed. The inside radius of the pulley rim is 100 mm and the centre of gravity of the shoe lies at 75 mm from the centre of the spider. Coefficient of friction may be taken as 0.3. Determine mass of the shoes.
a) 1.955 kg
b) 1.866 kg
c) 2.272 kg
d) 2.139 kg

Explanation: Given : P = 25 kW = 25 × 103 W ; N = 1200 r.p.m. or ω = 2 π × 1200/60 = 125.66 rad/s ; n = 4 ; R = 100 mm = 0.1 m ; r = 75 mm = 0.075 m ; µ = 0.3 ; ω1 = 0.5 ω.
ω1 = 0.5 ω = 0.5 x 125.66 = 62.83 rad/s
Power transmitted = Tω = T x 125.66
T = 198.95 N-m
Pc = mω2r = m x 125.662 x 0.075 = 1184.28 m N
Ps = mω12r = m x 62.832 x 0.075 = 296.07 m N
Frictional force acting tangentially on each shoe, F = µ(Pc–Ps) = 0.3 (1184.28 m – 296.07 m) = 266.46m N
Torque transmitted (T), 198.95 = n.F.R = 4 × 266.46 m × 0.1 = 106.584 m
m = 1.866 kg.

10. A centrifugal clutch is to transmit 22.5 kW at 1250 r.p.m. There are four shoes. The speed at which the engagement begins is 0.5 times the running speed. The inside radius of the pulley rim is 175 mm and the centre of gravity of the shoe lies at 120 mm from the centre of the spider. Coefficient of friction may be taken as 0.3. Determine the width of the shoe if the angle subtended by the shoes at the centre of the spider is 60° and the intensity of pressure exerted on the shoes is 0.05 N/mm2.
a) 104.493 mm
b) 89.198 mm
c) 52. 837 mm
d) 113.983 mm

Explanation: Given : P = 22.5 kW = 22.5 × 103 W ; N = 1250 r.p.m. or ω = 2 π × 1250/60 = 130.9 rad/s ; n = 4 ; R = 175 mm = 0.175 m ; r = 120 mm = 0.12 m ; µ = 0.3 ; ω1 = 0.5ω ; θ = 60° = π/3 rad ; p = 0.05 N/mm2
ω1 = 0.5ω = 0.5 x 130.9 = 65.45 rad/s
Power transmitted = Tω = T x 130.9
T = 171.887 N-m
Pc = mω2r = m x 130.92 x 0.12 = 2056.17 m N
Ps = mω12r = m x 65.452 x 0.12 = 514.04 m N
Frictional force acting tangentially on each shoe, F = µ(Pc– Ps) = 0.3 (2056.17 m – 514.04 m) = 462.64 m N
Torque transmitted (T), 171.887 = n.F.R = 4 × 462.64 m × 0.175 = 323.847 m
m = 0.53 kg
l = θ x R = π/3 x 175 = 183.26 mm
l.b.p = Pc – Ps = 1542.13 m
183.26 x b x 0.05 = 1542.13 x 0.53
b = 89.198 mm.

11. A centrifugal clutch has four shoes. When the clutch is at rest, each shoe is pulled against a stop by a spring so that it leaves a radial clearance of 10 mm between the shoe and the rim. The pull exerted by the spring is then 700 N. The mass centre of the shoe is 150 mm from the axis of the clutch. If the internal diameter of the rim is 500 mm, the mass of each shoe is 6 kg, the stiffness of each spring is 75 N/mm and the coefficient of friction between the shoe and the rim is 0.35; find the power transmitted by the clutch at 400 r.p.m.
a) 3.439 kW
b) 34.39 kW
c) 3439 kW
d) 3.439 W

Explanation: Given : n = 4 ; c = 10 mm ; S = 700 N ; r = 150 mm ; D = 500 mm or R = 250 mm = 0.25 m ; m = 6 kg ; s = 75 N/mm ; µ = 0.35 ; N = 400 r.p.m. or ω = 2 π × 400/60 = 41.89 rad/s
r1 = r + c = 150 + 10 = 160 mm = 0.16 m
Pc = m.ω2.r1 = 6 x (41.89)2 × 0.16 = 1684.58 N
Ps = S + c.s = 700 + 10 x 75 = 1450 N
F = µ (P¬c – Ps) = 0.35 (1684.58 – 1450) = 82.1 N
T = n.F.R = 4 × 82.1 × 0.25 = 82.1 N-m
Power transmitted (P) = Tω = 82.1 x 41.89 = 3439.31 W = 3.439 kW.

12. A centrifugal clutch has four shoes. When the clutch is at rest, each shoe is pulled against a stop by a spring so that it leaves a radial clearance of 7.5 mm between the shoe and the rim. The pull exerted by the spring is then 600 N. The mass centre of the shoe is 120 mm from the axis of the clutch. If the internal diameter of the rim is 350 mm, the mass of each shoe is 5 kg, the stiffness of each spring is 60 N/mm and the coefficient of friction between the shoe and the rim is 0.2; find the speed of the centrifugal clutch, if the power transmitted is 40.168 kW.
a) 83.779 r.p.m.
b) 800 r.p.m.
c) 104.66 r.p.m.
d) 1000 r.p.m.

Explanation: Given : n = 4 ; c = 7.5 mm ; S = 600 N ; r = 120 mm ; D = 350 mm or R = 175 mm = 0.175 m ; m = 5 kg ; s = 60 N/mm ; µ = 0.2; P = 40.168 kW = 40.168 x 103 W
r1 = r + c = 120 + 7.5 = 127.5 mm = 0.1275 m
Pc = m.ω2.r1 = 5 x (83.78)2 × 0.1275 = 4474.67 N
Ps = S + c.s = 600 + 7.5 x 60 = 1050 N
F = µ (Pc – Ps) = 0.2 (4474.67 – 1050) = 684.93 N
T = n.F.R = 4 × 684.93 × 0.175 = 479.45 N-m
Power transmitted (P) = Tω = 479.45 x ω
40.168 x 103 W = 479.45 x ω
Therefore, N = 60 x ω / (2 π) = 800 r.p.m.

13. A centrifugal clutch is to transmit 35 kW at 1500 r.p.m. There are four shoes. The speed at which the engagement begins is 0.7 times the running speed. The inside radius of the pulley rim is 120 mm and the centre of gravity of the shoe lies at 80 mm from the centre of the spider. Coefficient of friction may be taken as 0.35. Determine intensity of pressure exerted on the shoes if the angle subtended by the shoes at the centre of the spider is 60° and the width of the shoes is 105.52 mm.
a) 109.32 mm
b) 186.45 mm
c) 150.57 mm
d) 105.52 mm

Explanation: Given : P = 35 kW = 35 × 103 W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157.08 rad/s ; n = 4 ; R = 120 mm = 0.12 m ; r = 80 mm = 0.08 m ; µ = 0.35 ; ω1 = 0.7 ω ; θ = 60° = π/3 rad ; b = 105.52 mm. ω1 = 0.7 ω = 0.75 x 157.08 = 109.95 rad/s
Power transmitted = Tω = T x 104.72
T = 222.816 N-m
Pc = mω2r = m x 157.082 x 0.08 = 1973.93 m N
Ps = mω12r = m x 109.952 x 0.08 = 967.12 m N
Frictional force acting tangentially on each shoe, F = µ(Pc– Ps) = 0.35 (1973.93 m – 967.12 m) = 352.38 m N
Torque transmitted (T), 190.98 = n.F.R = 4 × 352.38 m × 0.12 = 169.144 m
m = 1.317 kg
l = θ x R = π/3 x 120 = 125.66 mm
l.b.p = Pc – Ps = 1006.81 m
125.66 x 105.52 x p = 1006.81 x 1.317
p = 0.1 N/mm2.

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