Theory of Machines Test 3

Result

Theory of Machines Test 3

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In a rim type flywheel, mass is distributed uniformly throughout. On a second rim type flywheel, mass is concentrated at 1/2 of the initial radius. If both wheels are rotating at the same angular velocity, what is the ratio of energy stored in the first flywheel to the second flywheel?

A
1/4

B
4

C
1/2

D
2

Solution

Concept:
\(E = \frac{1}{2}I{\omega ^2}\)
Calculation:
For first flywheel,
\({E_1} = \frac{1}{2}\left( {m{r^2}} \right){\omega ^2}\)
For second flywheel,
\({E_2} = \frac{1}{2}\left[ {m{{\left( {\frac{r}{2}} \right)}^2}} \right]{\omega ^2}\)
\({E_2} = \frac{1}{8}m{r^2}{\omega ^2}\)
Now,
\(\frac{{{E_1}}}{{{E_2}}} = \frac{{\frac{1}{2}\;m{r^2}{\omega ^2}}}{{\frac{1}{8}\;m{r^2}{\omega ^2}}}\)
\(\frac{{{E_1}}}{{{E_2}}} = 4\)
Question 2
1 / -0

A flywheel is fitted to the crank shaft of an engine having W amount of indicated work per revolution. Permissible limits of coefficient of fluctuation of energy and speed are C_E and C_S respectively. The kinetic energy of the flywheel is given by

A
\(2\frac{{W \cdot {C_E}}}{{{C_S}}}\)

B
\(\frac{{W \cdot {C_E}}}{{{2C_S}}}\)

C
\(\frac{{W \cdot {C_E}}}{{{C_S}}}\)

D
\(\frac{{W \cdot {C_s}}}{{{2C_E}}}\)

Solution

Concept:
Coefficient of fluctuation of energy:
\({C_E} = \frac{{Maximum\;fluctuation\;of\;energy}}{{Work\;done\;per\;cycle}}\)
\({C_E} = \frac{\Delta E}{W} \Rightarrow \Delta E=C_E.W\)
Coefficient of fluctuation of speed:
\(\begin{array}{l} {C_S} = \frac{{Maximum\;fluctuation\;of\;speed}}{{Mean\;speed}}\\ {C_S} = \frac{{{ω _1} - {ω _2}}}{ω } = \frac{{2\left( {{ω _1} - {ω _2}} \right)}}{{\left( {{ω _1} + {ω _2}} \right)}} \end{array}\)
ω = mean angular speed during the cycle (rad/s)
\(ω = \frac{{{ω _1} + {ω _2}}}{2}\)
Maximum fluctuation of energy: ΔE = CE. W
\(\begin{array}{l} {\rm{\Delta }}E = \frac{1}{2}Iω _1^2 - \frac{1}{2}Iω _2^2 = \frac{1}{2}I\left( {{ω _1} + {ω _2}} \right)\left( {{ω _1} - {ω _2}} \right)\\ {\rm{\Delta }}E = Iω \left( {{ω _1} - {ω _2}} \right) = I{ω ^2}\frac{{\left( {{ω _1} - {ω _2}} \right)}}{ω } = I{ω ^2}{C_S}\\ K.E = \frac{1}{2}I{ω ^2} = \frac{1}{2}.\frac{{{\rm{\Delta }}E}}{{{C_S}}} = \frac{{{C_E}W}}{{2{C_S}}} \end{array}\)
Question 3
1 / -0

If maximum and minimum equilibrium speed of governor are 700 rpm and 300 rpm respectively. What is the sensitivity of governor?

A
0.2

B
0.4

C
0.6

D
0.8

Solution

Concept:
\({\rm{Sensitivity\;of\;governor\;}} = \frac{{{\omega _{max}} - {\omega _{min}}}}{{{\omega _{mean}}}}\)
Calculation:
Given: ω_max = 700 rpm, ω_min = 300 rpm
Now,
\({\omega _{mean}} = \frac{{700 + 300}}{2} = 500\;rpm\)
\(\therefore {\rm{Sensitivity}} = \frac{{700 - 300}}{{500}}\)
∴ Sensitivity = 0.8
Question 4
1 / -0

For a porter governor the following data is obtained.
Mass of fly balls = 4 kg each, Mass of sleeve = 20 kg, Power = 4 J, Maximum rise of sleeve = 250 mm
For the information given above, calculate the fraction change in speed. (round up to two decimal places)

A
0-1

B
04-11

C
07-16

D
09-12

Solution

Concept:
Power of generator is given as
\(P = \left( {m + M} \right)gh\;\left( {\frac{{4{c^2}}}{{1\; + \;2c}}} \right)\)
Calculation:
Given:
Mass of fly balls (m) = 4 kg, Mass of sleeve (M) = 20 kg, Maximum rise of sleeve (h) = 250 mm = 0.25 m, Power = 4 J
\(P = \left( {m + M} \right)gh\;\left( {\frac{{4{c^2}}}{{1\; + \;2c}}} \right)\)
\(4 = \left( {4 + 20} \right)9.81 \times 0.25 \times \frac{{4{c^2}}}{{1\; + \;2c}}\)
\( \Rightarrow 4 + 8c = 24 \times 9.81 \times \frac{1}{4} \times 4{c^2}\)
4 + 8c = 24 × 9.81 c²
⇒ 58.86c² – 2c – 1 = 0
∴ c = 0.1484
Question 5
1 / -0

Torque requirement of a punching machine is given as T= 200 + 30sinθ, where θ is a crank angle. A flywheel with 15 kg.m² moment of inertia is attached. If power is supplied at a uniform rate, what is the maximum acceleration of flywheel possible (rad/s²)?

A
1

B
2

C
3

D
4

Solution

Concept:
T_max = Iα_max
Calculation:
T = 200 + 30 sin θ
\({T_{mean}} = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } \left( {200 + 30\sin \theta } \right)d\theta \)
\({T_{mean}} = \frac{1}{{2\pi }}\left[ {200\;\theta + 30\;\left( { - \cos \theta } \right)} \right]_0^{2\pi }\)
∴ T_mean = 200 N.m
Now,
T_flywheel = T - T_mean
T_flywheel = 200 + 30 sin θ – 200
T_flywheel = 30 sin θ
We know that, maximum value of sin θ = 1
∴ T_max = T_flywheel = 30 N.m
T_max = Iα_max
\(\therefore {\alpha _{max}} = \frac{{30}}{{15}}\)
∴ α_max= 2 rad/s²
Question 6
1 / -0

A steam engine develops 300 kW power at 9.5 rad/s. The coefficient of fluctuation of energy is found to be 0.1 and fluctuation of speed is kept within ± 0.5% of mean speed. Find the mass of the flywheel (in kg) required if the radius of gyration is 2 m.

A
5494-5500

B
54944-55001

C
54947-55006

D
54949-55002

Solution

Concept:
Coefficient of fluctuation of energy:
\({C_E} = \frac{{Maximum\;Fluctuation\;of\;energy}}{{Work\;done\;per\;cycle}}\)
Work done per cycle:
W = T_mean × θ
\({T_{mean}} = \frac{{P \times 60}}{{2\pi N}} = \frac{P}{\omega }\)
Work done per cycle:
\(W = \frac{{P \times 60}}{n}\)
θ = 2π (in case of steam engine and two stroke IC engine)
θ = 4π (in case of four stroke IC engine)
Coefficient of fluctuation of speed:
\({C_s} = \frac{{{\omega _{max}} - {\omega _{min}}}}{{{\omega _{mean}}}} = \frac{{2\left( {{\omega _1} - {\omega _2}} \right)}}{{{\omega _1} + {\omega _2}}}\)
Mean kinetic energy of flywheel:
\(E = \frac{1}{2}I{\omega ^2} = \frac{1}{2}m{k^2}{\omega ^2}\)
As the speed of the flywheel changes from ω₁ to ω₂, the maximum fluctuation of energy:
ΔE = 2EC_s
Calculation:
\({C_s} = \frac{{1.005\;\omega - 0.995\;\omega }}{\omega } = 0.01\)
\(W = \frac{{P \times 60}}{N} = \frac{P}{\omega } \times 2\pi \)
\(W = \frac{{300}}{{9.5}} \times 2\pi = 198.42\;kN.m\)
The maximum fluctuation of energy:
ΔE = W × C_E = 198.42 × 0.1 = 19.84 kN.m
ΔE = 2EC_s = mk²ω²C_s
\(m = \frac{{{\rm{\Delta }}E}}{{{k^2}{\omega ^2}{C_s}}} = \frac{{19.84 \times {{10}^3}}}{{{{\left( 2 \right)}^2} \times {{\left( {9.5} \right)}^2} \times 0.01}}\)
∴ m = 5495.84 kg
Question 7
1 / -0

In a cam design, the rise motion is given by a simple harmonic motion (SHM)
\(s = \frac{h}{2}\left( {1 - cos\frac{{\pi \theta }}{\beta }} \right)\). Where, h is total rise, θ is camshaft angle, β is the total angle of the rise interval.
For, h = 20 mm, speed = 300 rpm and β = 60˚, the maximum velocity (m/s) is given as ____

A
0.93-0.95

B
0.934-0.951

C
0.937-0.956

D
0.939-0.952

Solution

Concept:
\(V = \omega \frac{{ds}}{{d\theta }}\)
Calculation:
Given: h = 20 mm, speed = 300 rpm and β = 60˚
Now,
\(s = \frac{h}{2}\left( {1 - \cos \frac{{\pi \theta }}{\beta }\;} \right)\)
\(\therefore \frac{{ds}}{{d\theta }} = \frac{h}{2}(0 + \sin \frac{{\pi \theta }}{\beta })\frac{\pi }{\beta }\)
\(\frac{{ds}}{{d\theta }} \Rightarrow \max ,{\rm{\;}}\;when{\rm{\;sin}}\frac{{\pi \theta }}{\beta } = 1\)
\(\therefore {\left( {\frac{{ds}}{{d\theta }}} \right)_{max}} = \frac{{h\pi }}{{2\beta }}\)
h = 20 mm = 0.02 m
\(\omega = \frac{{2\pi \times 300}}{{60}} = 31.41\;rad/s\)
\(\beta = 60^\circ = \frac{\pi }{3}\)
\({V_{max}} = \frac{{0.02 \times 31.41 \times \pi }}{{2 \times \left( {\frac{\pi }{3}} \right)}}\)
∴ V_max = 0.9423 m/s
Question 8
1 / -0

The cross-section of a flywheel is square and the rim has a mean radius of 0.50 m. The density of flywheel is 7000 kg/m³ and it is employed in a punching press which runs at 240 rpm. During punching operation which lasts one-quarter of a revolution energy requirement from flywheel is 3 kJ. If the coefficient of speed fluctuation is 0.18 and the rim contributes 85% of total moment of inertia then the cross section of the flywheel is _________× 10^-3 m².

A
3.9-4.2

B
3.94-4.21

C
3.97-4.26

D
3.99-4.22

Solution

Concept:
The energy required by flywheel is \(\Delta E\)
Contribution of rim to Moment Of Inertia = k
Coefficient of speed fluctuation = C_s
\({\rm{Angular\;velocity\;}}\left( {\rm{\omega }} \right){\rm{\;}} = \frac{{2\pi N}}{{60}}\;rad/s\)
\(\therefore k\Delta E = I{\omega ^2}{C_s}\)
Where, I = moment of inertia = mass × (mean radius)²
Calculation:
ΔE = 3000 J
Contribution of rim, k = 85%
k = 0.85, C_s = 0.18, N= 240 rpm, density (ρ) = 7000 kg/m²
\(\omega = \frac{{2\pi N}}{{60}} = 8\pi \;rad/s\)
kΔE = Iω² C_s
\(I = \frac{{0.85\; × \; 3000}}{{64{\pi ^2}\; × \; 0.18}}\)
∴ I = 22.427 kg.m²
\(m = \frac{I}{{{R^2}}} = \frac{{22.427}}{{0.25}}\)
∴ m = 89.711 kg
Mass, m = (2πR) × bt × ρ (∵ mass = denity × volume)
where bt = cross-sectional area
\(bt = \frac{{89.711}}{{2\pi\; × \;0.5\; × \;7000}}\)
∴ bt = 4.079 × 10^-3 m²
Question 9
1 / -0

In a spring controlled governor, controlling force varies linearly with radius of rotation. When controlling force is 1500 N, radius of rotation is 200 mm and when controlling force is 800 N, radius of rotation is 120 mm. By changing initial tension governor behaves as isochronous governor. What is the value of change in tension and isochronous speed? (mass of balls = 8 kg)

A
250N, 316rpm

B
316N, 250rpm

C
250N, 33.07rpm

D
316N, 33.07rpm

Solution

Explanation:
F = ar + b, a and b are constants
1500 = 200 a + b
800 = 120 a + b
Solving, we get
a = 8.75 N/mm, b = -250 N
for isochronous governor, b = 0
∴ F = ar
F = mr ω² = ar
mω² = a
\(\omega = \sqrt {\frac{a}{m}} \)
\(\omega = \sqrt {\frac{{8.75 \times 1000\left( {N/m} \right)}}{8}} \)
ω = 33.07 rad/s
\(N = \frac{{\omega \times 60}}{{2\pi }}\)
∴ N = 315.8 rpm ≈ 316 rpm
Now,
Change in tension = 0 – (-250)
∴ Change in tension = 250 N

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Theory of Machines Test 3

Result

Theory of Machines Test 3

Score

-

Rank

-

SHARING IS CARING

Question 1

1/4

4

1/2

2

Solution

Question 2

\(2\frac{{W \cdot {C_E}}}{{{C_S}}}\)

\(\frac{{W \cdot {C_E}}}{{{2C_S}}}\)

\(\frac{{W \cdot {C_E}}}{{{C_S}}}\)

\(\frac{{W \cdot {C_s}}}{{{2C_E}}}\)

Solution

Question 3

0.2

0.4

0.6

0.8

Solution

Question 4

0-1

04-11

07-16

09-12

Solution

Question 5

1

2

3

4

Solution

Question 6

5494-5500

54944-55001

54947-55006

54949-55002

Solution

Question 7

0.93-0.95

0.934-0.951

0.937-0.956

0.939-0.952

Solution

Question 8

3.9-4.2

3.94-4.21

3.97-4.26

3.99-4.22

Solution

Question 9

250N, 316rpm

316N, 250rpm

250N, 33.07rpm

316N, 33.07rpm

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.