Explanation The following figure represents the machine component with uniform diameter of the aluminum wire. Figure-(1) Write the expression for the length of the component 1. L 1 = π r 1 2 Here, the length of the component 1 is L 1 and the radius of the component 1 is r 1 . Write the expression for the length of the component 2. L 2 = r 2 − r 1 Here, the length of the component 2 is L 2 and the radius of the component 2 is r 2 . Write the expression for the length of the component 3. L 3 = π r 2 2 Here, the length of the component 3 is L 3 . Write the expression for the length of the component 4. L 4 = r 2 − r 1 Here, the length of the component 4 is L 4 . Write the expression for the length of the component 5. L 5 = π r 1 2 Here, the length of the component 5 is L 5 . Write the expression for the mass of component 1. m 1 = m L 1 ........ (I) Here, the mass of the component 1 is m 1 , the length of the component 1 is L 1 and the mass per unit length of the wire is m . Write the expression for the mass of component 2. m 2 = m L 2 ....... (II) Here, the mass of the component 2 is m 2 and the length of the component 2 is L 2 . Write the expression for the mass of component 3. m 3 = m L 3 ....... (III) Here, the mass of the component 3 is m 3 and the length of the component 3 is L 3 . Write the expression for the mass of component 4. m 4 = m L 4 ....... (IV) Here, the mass of the component 4 is m 4 and the length of the component 4 is L 4 . Write the expression for the mass of component 4. m 5 = m L 5 ....... (V) Here, the mass of the component 5 is m 5 and the length of the component 5 is L 5 . Substitute π r 1 2 for L 1 in Equation (I). m 1 = π r 1 2 ⋅ m = m π r 1 2 Substitute r 2 − r 1 for L 2 in Equation (II). m 2 = m ( r 2 − r 1 ) Substitute π r 2 2 for L 3 in Equation (III). m 3 = π r 2 2 ⋅ m = m π r 2 2 Substitute r 2 − r 1 for L 4 in Equation (IV). m 4 = ( r 2 − r 1 ) ⋅ m = m ( r 2 − r 1 ) Substitute π r 1 2 for L 5 in Equation (V). m 5 = π r 1 2 ⋅ m = m π r 1 2 All the mass product of inertia from the centroidal axes is equal to zero because of symmetry. Write the expression for the x coordinates of the center of gravity of the component 1. x ¯ 1 = 0 Write the expression for the x coordinates of the center of gravity of the component 2. x ¯ 2 = 0 Write the expression for the x coordinates of the center of gravity of the component 3. x ¯ 3 = x ¯ 3 Write the expression for the x coordinates of the center of gravity of the component 4. x ¯ 4 = x ¯ 4 Write the expression for the y coordinates of the center of gravity of the component 1. y ¯ 1 = y ¯ 1 Write the expression for the y coordinates of the center of gravity of the component 2. y ¯ 2 = 0 Write the expression for the y coordinates of the center of gravity of the component 3. y ¯ 3 = 0 Write the expression for the y coordinates of the center of gravity of the component 4. y ¯ 4 = 0 Write the expression for the y coordinates of the center of gravity of the component 5. y ¯ 5 = y ¯ 5 Write the expression for the z coordinates of the center of gravity of the component 5. z ¯ 5 = 0 Write the expression for the z coordinates of the center of gravity of the component 4. z ¯ 4 = 0 Write the expression for the mass product of inertia of the element with respect to u and v axes. d I u v = u ⋅ v ⋅ d m ........ (VI) Here, the mass product of the inertia of the element with respect to u and v axes is d I u v , the horizontal distance of the element is u and the vertical distance of the element is v . Write the expression for the horizontal distance of the element. u = a ⋅ cos θ Here, the radius of the element is a and the angle is θ . Write the expression for the vertical distance of the element. v = a ⋅ sin θ Write the expression for the mass of the element. d m = ρ d V ....... (VII) Here, the density of the element is ρ , the volume of the element is d V and the mass of the element is d m . Write the expression for the volume of the element. d V = A ⋅ a ⋅ d θ Here, the area of the element is A and the length of the element is a ⋅ d θ . Write the expression for the mass per unit length of the element. m = ρ A A = m ρ ....... (VIII) Substitute a ⋅ sin θ for v and a ⋅ cos θ for u in Equation (VI). d I u v = ( a ⋅ cos θ ) ⋅ ( a ⋅ sin θ ) ⋅ d m d I u v = a 2 ⋅ cos θ ⋅ sin θ ⋅ d m ........ (IX) Substitute A ⋅ a ⋅ d θ for d V in Equation (VII). d m = ρ ⋅ A ⋅ a ⋅ d θ Substitute ρ ⋅ A ⋅ a ⋅ d θ fore d m in Equation (IX). d I u v = a 2 ⋅ cos θ ⋅ sin θ ⋅ ρ ⋅ A ⋅ a ⋅ d θ d I u v = ρ ⋅ A ⋅ a 3 ⋅ cos θ ⋅ sin θ ⋅ d θ ....... (X) Substitute m ρ for A in Equation (X). d I u v = ρ ⋅ m ρ ⋅ a 3 ⋅ cos θ ⋅ sin θ ⋅ d θ = m ⋅ a 3 ⋅ cos θ ⋅ sin θ ⋅ d θ ....... (XI) Integrate Equation (XI) with respect to θ ∫ 0 I u v d I u v = ∫ 0 π / 2 m ⋅ a 3 ⋅ cos θ ⋅ sin θ ⋅ d θ [ I u v ] 0 I u v = m ⋅ a 3 ∫ 0 π / 2 cos θ ⋅ sin θ ⋅ d θ I u v = m ⋅ a 3 2 ∫ 0 π / 2 2 cos θ ⋅ sin θ ⋅ d θ I u v = m ⋅ a 3 2 [ sin 2 θ ] 0 π / 2 I u v = m ⋅ a 3 2 [ ( sin ( π / 2 ) ) 2 − sin 0 ° ] I u v = m ⋅ a 3 2 ⋅ 1 I u v = m ⋅ a 3 2 ....... (XII) Write the expression for the mass of products of the inertia to the centroidal axes for component 1. ( I ¯ y ′ z ′ ) 1 = m r 1 3 2 Write the expression for the mass of products of inertia to the centroidal axes for component 3. ( I ¯ x ′ z ′ ) 3 = − m r 2 3 2 Write the expression for the mass of products of the inertia to the centroidal axes for component 5. ( I ¯ y ′ x ′ ) 5 = − m r 1 3 2 The following table represents the required values of each component of formed aluminum wire. m x ¯ y ¯ z ¯ m x ¯ ⋅ y ¯ m z ¯ ⋅ y ¯ m z ¯ ⋅ x ¯ m π r 1 2 0 y ¯ 1 0 0 0 0 m ( r 2 − r 1 ) 0 0 0 0 0 0 m π r 2 2 x ¯ 3 0 0 0 0 0 m ( r 2 − r 1 ) x ¯ 4 0 0 0 0 0 m π r 1 2 0 y ¯ 5 0 0 0 0 Write the expression for the mass product of the inertia with respect to x and y axes

12th Edition

ISBN: 9781259977237

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

Videos

Question

Chapter B, Problem B.48P

To determine

The product of mass moment of inertia Izx.

The product of mass moment of inertia Ixy.

The product of mass moment of inertia Iyz.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter B, Problem B.47P

Chapter B, Problem B.49P

Students have asked these similar questions

thank you for previous answer I apologize if the acceleration was unclear it is underlined now along with values in tables

११११११११ TABLE Much 160,000kg Croll 0,005 CD Ap Par ng При nchs 0.15 5m² 1.2kg/m³ 0.98 0.9 0,98 0,9 0,88 IF 20 10 to add The train is going to make several stops along its journey. It will be important for the train to accelerate quickdy to get back up to speed. In order to get Tesla Model S motors until we get the combined The Forque and power needed we are goins bined power and forque needed to accelerate from 0 to 324 km/hr in less than 5 Minutes. Tesla Prated 270 kW Tesla Trated Twheel ng Jaxle 440 NM 20 8.5kgm² 0.45M a) What is the minimum whole number of Tesla Motors required to achieve accelerate the train from 0 to 324 km/hr in less than 5 Nnutes? Seperate the acceleration into constant torque and constant power 0. b) How long does it take the train to accelerate from 0 to 324 km/hr with the number of Tesla motors from part a? c) Using Matlab plot the relocity profile as a function of time, Is this a constant acceleration profile? B

Example find f(t)? -4s F(s)= (s² + 4)²

Chapter B Solutions

Vector Mechanics For Engineers

Ch. B - A thin plate with a mass m is cut in the shape of...Ch. B - Prob. B.2P Ch. B - Prob. B.3P Ch. B - Prob. B.4P Ch. B - A piece of thin, uniform sheet metal is cut to...Ch. B - Prob. B.6P Ch. B - Prob. B.7P Ch. B - Prob. B.8P Ch. B - Prob. B.9P Ch. B - Prob. B.10P

Ch. B - Prob. B.11P Ch. B - Prob. B.12P Ch. B - Determine by direct integration the mass moment of...Ch. B - Prob. B.14P Ch. B - A thin, rectangular plate with a mass m is welded...Ch. B - A thin steel wire is bent into the shape shown....Ch. B - Prob. B.17P Ch. B - Prob. B.18P Ch. B - Prob. B.19P Ch. B - Prob. B.20P Ch. B - Prob. B.21P Ch. B - Prob. B.22P Ch. B - Prob. B.23P Ch. B - Prob. B.24P Ch. B - Prob. B.25P Ch. B - Prob. B.26P Ch. B - Prob. B.27P Ch. B - Prob. B.28P Ch. B - Prob. B.29P Ch. B - Prob. B.30P Ch. B - Prob. B.31P Ch. B - Determine the mass moments of inertia and the...Ch. B - Prob. B.33P Ch. B - Prob. B.34P Ch. B - Prob. B.35P Ch. B - Prob. B.36P Ch. B - Prob. B.37P Ch. B - Prob. B.38P Ch. B - Prob. B.39P Ch. B - Prob. B.40P Ch. B - Prob. B.41P Ch. B - Prob. B.42P Ch. B - Prob. B.43P Ch. B - Prob. B.44P Ch. B - A section of sheet steel 2 mm thick is cut and...Ch. B - Prob. B.46P Ch. B - Prob. B.47P Ch. B - Prob. B.48P Ch. B - Prob. B.49P Ch. B - Prob. B.50P Ch. B - Prob. B.51P Ch. B - Prob. B.52P Ch. B - Prob. B.53P Ch. B - Prob. B.54P Ch. B - Prob. B.55P Ch. B - Determine the mass moment ofinertia of the steel...Ch. B - Prob. B.57P Ch. B - Prob. B.58P Ch. B - Determine the mass moment of inertia of the...Ch. B - Prob. B.60P Ch. B - Prob. B.61P Ch. B - Prob. B.62P Ch. B - Prob. B.63P Ch. B - Prob. B.64P Ch. B - Prob. B.65P Ch. B - Prob. B.66P Ch. B - Prob. B.67P Ch. B - Prob. B.68P Ch. B - Prob. B.69P Ch. B - Prob. B.70P Ch. B - For the component described in the problem...Ch. B - Prob. B.72P Ch. B - For the component described in the problem...Ch. B - Prob. B.74P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The product of mass moment of inertia I z x . The product of mass moment of inertia I x y . The product of mass moment of inertia I y z .

Solution Summary: The following figure represents the machine component with uniform diameter of the aluminum wire.

Concept explainers

Videos

The product of mass moment of inertia Izx.

The product of mass moment of inertia Ixy.

The product of mass moment of inertia Iyz.

Want to see the full answer?

Chapter B Solutions

The product of mass moment of inertia I z x . The product of mass moment of inertia I x y . The product of mass moment of inertia I y z .

Solution Summary: The following figure represents the machine component with uniform diameter of the aluminum wire.

Concept explainers

Videos

The product of mass moment of inertia Izx. The product of mass moment of inertia Ixy. The product of mass moment of inertia Iyz.

Want to see the full answer?

Chapter B Solutions

The product of mass moment of inertia Izx.

The product of mass moment of inertia Ixy.

The product of mass moment of inertia Iyz.