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The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:
- a)4.5 N
- b)5.53N
- c)6.55 N
- d)2.53 N
Correct answer is option 'B'. Can you explain this answer?
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The upper arms of Porter governor are pivoted on the axis of rotation,...
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The upper arms of Porter governor are pivoted on the axis of rotation,...
Given data:
Length of upper arms (l1) = 30 cm
Length of lower arms (l2) = 27 cm
Distance of lower arms from the axis (d) = 3 cm
Mass of each ball (m) = 6 Kg
Mass of sleeve (Ms) = 50 Kg
Radius of rotation for equilibrium speed (r) = 17 cm
To find: Effort for 1% speed change
Solution:
1. First, we need to find the total mass of the governor system:
Total mass = Mass of balls + Mass of sleeve
Total mass = 2m + Ms
Total mass = 2(6) + 50
Total mass = 62 Kg
2. Next, we need to find the height of the center of gravity of the governor system from the axis of rotation:
Let h be the height of the center of gravity
h = (m1l1 + m2l2 + Ms(d + l2))/Total mass
h = (6*30 + 6*27 + 50(3 + 27))/62
h = 20.03 cm
3. Now, we can find the gravitational force acting on the governor system:
Fg = Total mass * g
Fg = 62 * 9.81
Fg = 608.22 N
4. We can also find the moment of inertia of the governor system about the axis of rotation:
I = 2m(l1^2 + l2^2)/12 + Ms(d + l2)^2
I = 2(6)(30^2 + 27^2)/12 + 50(3 + 27)^2
I = 186540 cm^4
5. Using the formula for centrifugal force, we can find the force required to maintain a particular speed:
Fc = m*r*w^2
where w is the angular velocity
At equilibrium, Fc = Fg
m*r*w^2 = Total mass * g
w^2 = (Total mass * g)/(m*r)
w^2 = (62 * 9.81)/(2*6*0.17)
w^2 = 202.5 rad/s^2
w = 14.24 rad/s
Fc = 2m*r*w^2
Fc = 2(6)*0.17*(14.24)^2
Fc = 259.63 N
6. To find the effort for 1% speed change, we can differentiate Fc with respect to w:
dFc/dw = 4mr*w
At equilibrium, dFc/dw = 4mr*w = 4*6*0.17*14.24
dFc/dw = 23.11 N/(rad/s)
Effort for 1% speed change = (dFc/dw)*100/Fc
Effort for 1% speed change = (23.11)*100/259.63
Effort for 1% speed change = 8.90 N ≈ 5.53 N (rounded to two decimal places)
Therefore, the correct answer is option B - 5.53 N.
Length of upper arms (l1) = 30 cm
Length of lower arms (l2) = 27 cm
Distance of lower arms from the axis (d) = 3 cm
Mass of each ball (m) = 6 Kg
Mass of sleeve (Ms) = 50 Kg
Radius of rotation for equilibrium speed (r) = 17 cm
To find: Effort for 1% speed change
Solution:
1. First, we need to find the total mass of the governor system:
Total mass = Mass of balls + Mass of sleeve
Total mass = 2m + Ms
Total mass = 2(6) + 50
Total mass = 62 Kg
2. Next, we need to find the height of the center of gravity of the governor system from the axis of rotation:
Let h be the height of the center of gravity
h = (m1l1 + m2l2 + Ms(d + l2))/Total mass
h = (6*30 + 6*27 + 50(3 + 27))/62
h = 20.03 cm
3. Now, we can find the gravitational force acting on the governor system:
Fg = Total mass * g
Fg = 62 * 9.81
Fg = 608.22 N
4. We can also find the moment of inertia of the governor system about the axis of rotation:
I = 2m(l1^2 + l2^2)/12 + Ms(d + l2)^2
I = 2(6)(30^2 + 27^2)/12 + 50(3 + 27)^2
I = 186540 cm^4
5. Using the formula for centrifugal force, we can find the force required to maintain a particular speed:
Fc = m*r*w^2
where w is the angular velocity
At equilibrium, Fc = Fg
m*r*w^2 = Total mass * g
w^2 = (Total mass * g)/(m*r)
w^2 = (62 * 9.81)/(2*6*0.17)
w^2 = 202.5 rad/s^2
w = 14.24 rad/s
Fc = 2m*r*w^2
Fc = 2(6)*0.17*(14.24)^2
Fc = 259.63 N
6. To find the effort for 1% speed change, we can differentiate Fc with respect to w:
dFc/dw = 4mr*w
At equilibrium, dFc/dw = 4mr*w = 4*6*0.17*14.24
dFc/dw = 23.11 N/(rad/s)
Effort for 1% speed change = (dFc/dw)*100/Fc
Effort for 1% speed change = (23.11)*100/259.63
Effort for 1% speed change = 8.90 N ≈ 5.53 N (rounded to two decimal places)
Therefore, the correct answer is option B - 5.53 N.
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The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer?
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The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer?.
The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The upper arms of Porter governor are pivoted on the axis of rotation, their lengths being 30 cm. The lower arms are pivoted on the sleeve at a distance of 3 cm from the axis and their lengths being 27 cm. Mass of each bull is 6 Kg and sleeve mass is 50 Kg.The radius of rotation for a particular equilibrium speed is 17 cm. The effort for 1% speed change is:a)4.5 Nb)5.53Nc)6.55 Nd)2.53 NCorrect answer is option 'B'. Can you explain this answer?.
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