The mass center, G, of the 2-lb ball has a velocity of (vd - 5 ft/s when it strikes theend of the smooth 4-lb slender bar, which is at rest. The bar has a length l-Sft.The coefficient of restitution is e-0.8.The goal is to find the angular velocity of the bar after impact.Steo 5: Finally find the angular velocity of the bar after impact in rad/s. Give youranswer with 2 decimal places

Here we have mass of ball =2lb(pound)initial velocity of ball=5ft/slength of bar =5ftHeight of bar =0.5ftcoefficient of restituition (e) =0.8Mass of bar =4lb(Pound)solutionaccording to the law of conservation of momentum M1U1=M2U2=M1V1=M2V2................(1)→1lb(pound)=0.454 kgMass of ball =0.908 kgMass of bar =1.816kg→ 1ft/s=0.305 meter(u1)BALL=1.525 m/snow putting the value in eqn (1)(0.908 x 1.525)+(1.816 x 0)=(0.908 x V1)+(1.816 X V2)Initially bar is at rest so u2=00.908V1+1.816V2=1.3847...........(2)We know that V2-V1=e(u1-u2)v2-v1=0.8 x 1.525v2-v1=1.22v2=(1.22 + V1)...........................(3)putting the values of v2 in eqn (2)0.908 V1+1.816(1.22 + V1)=1.38470.908V1+1.816V1+2.2155=1.38472.724V1=0.8308V1=0.3049VELOCITIES ARE ALWAYS POSITIVE so v1=0.3049 m/snow from eqn (3)v2=1.22+0.3049v2=1.5249 m/snoe the angular velocity of the bar is V=rww=Vr (r=l2=2.5ft)=v2r=1.5249(2.5 x 0.305) (converting feet to meter )=1.52490.7625Wbar=1.999 rad/seclinear velocity of the bar at point B is,V2=1.5249m/sV2=4.999 ft/s→Moment of inertia of the barby the parallel axis theorem I0=IG+M(OG)2 =IG+M(x)2 =ML212+M(x)2 as M=4.2lbas 1 Pound =0.0311 slugso 4 pound =0.1244 slugID=(0.1244) X (5)212+ (0.1244) X (0.5)2 =0.1244 X 2512+(0.1244 X 0.25)=0.259 +0.031ID=0.290 slug ft2Relation between Vbar & VballM1U1+M2U2 =M1V1+M2V2(U2=0)M1U1+M2U2=M1V1+M2V2M1U1=M1VBALL +M2Vbar2x5=2VBALL+4VBAR10=2VBALL+4Vbar...................................(1)dividing equation 1 by (12.8865)1012.8865=2Vball12.8865+4VBAR12.88650.776=0.155VBALL+0.310Vbar0.908 vball+1.816Vbar=1.3847(in kg and meter)relation of velocity of bar and ball as (v2-v1)=e(u1-u2)(v2-v1)=0.8 x 5v2-v1=4Vbar-Vball=4

Engineering

Mechanical Engineering

The mass center, G, of the 2-lb ball has a velocity of Ivch - 5 tis when it strikes the end of the smooth 4-b slender bar, which is at rest. The bar has a length l-St. The coefficient of restitution is e-0.8. B The goal is to find the angular velocity of the bar after impact. Stee 5: Finally find the angular velocity of the bar after impact in rad/s. Give your answer with 2 decimal places

The mass center, G, of the 2-lb ball has a velocity of Ivch - 5 tis when it strikes the end of the smooth 4-b slender bar, which is at rest. The bar has a length l-St. The coefficient of restitution is e-0.8. B The goal is to find the angular velocity of the bar after impact. Stee 5: Finally find the angular velocity of the bar after impact in rad/s. Give your answer with 2 decimal places

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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