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King Fahd University of Petroleum & Minerals

King Fahd University of Petroleum & Minerals. Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 41. Conservation of Momentum. Conservation of Linear Momentum. Conservation of Angular Momentum. Example 19-7. Example 19-7. Example 19-7. Impact.

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King Fahd University of Petroleum & Minerals

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  1. King Fahd University of Petroleum & Minerals Mechanical Engineering Dynamics ME 201 BY Dr. Meyassar N. Al-Haddad Lecture # 41

  2. Conservation of Momentum • Conservation of Linear Momentum • Conservation of Angular Momentum

  3. Example 19-7

  4. Example 19-7

  5. Example 19-7

  6. Impact • Impact occurs when two bodies collide with each other during a very short period of time. • Types of impact: • Central impact • Oblique impact Line of impact Plane of impact

  7. Coefficient of restitution e Coefficient of restitution e is defined as the ratio of the restitution impulse To the deformation impulse. Coefficient of restitution e is defined as the ratio of relative velocity after impact to the relative velocity before impact Coefficient of restitution e range between 0-1 Elastic impact e = 1(re-bounce with same velocity) Plastic impact e = 0 (couple or stick together and move with common velocity)

  8. Procedure for Analysis • Identify the intial velocity “ “you may use” • T1+ V1 = T2+ V2 • Apply the conservation of momentum along the line of impact, you will get one equation with two unknown velocity • Use the coefficient of restitution to obtain a second equation • Solve both equation for final velocities after the impact

  9. Eccentric Impact Line of the impact does not coincide with the mass centers of the two bodies One or both of the bodies are constrained to rotate about a fixed axis Angular momentum is conserved at point of rotation

  10. Procedure for Analysis • Apply the conservation of angular momentum at a rotational point, you will get one equation with two unknown ( v and w ) • Use the coefficient of restitution to obtain a second equation O A B

  11. Example 19-8 WR=10 Ib WB=2-Ib (vB)1= 30 ft/s wR= ? e = 0.4

  12. Problem 19-40 mp= 7 g vP= 800 m/s mD= 5 kg wD= ? q = ? stop

  13. Problem 19-49 W= 6 Ib From rest at horizontal Wbl= 1 Ib Vbl= 50 ft/s wrod=? e= 0.7 1 2

  14. Thank you

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