Explanation Given information: The moment of inertia ( I A A ′ ) with respect to A A ′ is 2 , 000 in 4 . The moment of inertia ( I B B ′ ) with respect to B B ′ is 4 , 000 in 4 . The distance ( d 1 ) from the center to the axis A A ′ is 8 in . The distance B ( d 2 ) from the center to axis B B ′ is 4 in . Calculation: Find the shaded area ( A ) using parallel axis theorem as shown below: I A A ′ = I ¯ + A d 1 2 (1) Substitute 8 in . for ( d 1 ) and 2 , 000 in 4 for I A A ′ in Equation (1). 2000 = I ¯ + A ( 8 ) 2 2000 = I ¯ + A ( 64 ) (2) Applying parallel axis theorem as shown below: I B B ′ = I ¯ + A ( d 1 + d 2 ) 2 (3) Substitute 4 in . for ( d 2 ) 8 in . for ( d 1 ) , and 4 , 000 in 4 for I B B ′ in Equation (3)

12th Edition

ISBN: 9781260663778

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 9.2, Problem 9.37P

To determine

Find the shaded area and its moment of inertia with respect to the centroidal axis parallel to AA′.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.2, Problem 9.36P

Chapter 9.2, Problem 9.38P

Students have asked these similar questions

The wall of a furnace has a thickness of 5 cm and thermal conductivity of 0.7 W/m-°C. The inside surface is heated by convection with a hot gas at 402°C and a heat transfer coefficient of 37 W/m²-°C. The outside surface has an emissivity of 0.8 and is exposed to air at 27°C with a heat transfer coefficient of 20 W/m²-ºC. Assume that the furnace is inside a large room with walls, floor and ceiling at 27°C. Show the thermal circuit and determine the heat flux through the furnace wall. h₁ T₁ k -L T. sur ho E

Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ... title('Velocity Triangle'), ... xlabel('x'),ylabel('y'), grid

To save fuel during the heating season it is suggested that glass windows be covered at night with a 1.2 cm layer of polystyrene. Estimate the percent savings in energy and discuss the feasibility of this idea. Show the thermal circuit with and without the insulation panel. Consider a typical case of 0.2 cm thick window glass with inside and outside heat transfer coefficients of 6 and 32 W/m²-ºC. Lg←←Lp h T₁ T。 g kp insulation panel

Chapter 9 Solutions

VECTOR MECH. FOR EGR: STATS & DYNAM (LL

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.

Find the shaded area and its moment of inertia with respect to the centroidal axis parallel to A A ′ .

Solution Summary: The author calculates the shaded area and its moment of inertia with respect to AAprime .

Concept explainers

Videos

Find the shaded area and its moment of inertia with respect to the centroidal axis parallel to AA′.

Want to see the full answer?

Chapter 9 Solutions