Mechanics - First Year Engineering, Study notes of Mechanics

Download Mechanics - First Year Engineering and more Study notes Mechanics in PDF only on Docsity! Peoples Empowerment Group ISBM COLLEGE OF ENGINEERING, NANDE, PUNE DEPARTMENT OF APPLIED SCIENCE Academic Year 2019-20-21 ENGINEERING MECHANICS Experiment no. 6 Title: To find coefficient of restitution, impulse of rate of energy loss for a direct central impact between the two bodies. Aim: To find coefficient of restitution, impulse of rate of energy loss for a direct central impact between the two bodies. Apparatus: A rubber ball, Meter scale, plastic ball, marble ball Theory: When a particle of mass ‘m’ moving with velocity v is called upon by force ‘f’ Newton’s law F= d dt (mv) or at F.dt = d(mv) . Integrating this for time t1 to time t2 , when the velocity change from v1 to v2 the equation becomes mv1 = ∫ f dt= mv2. The integral in equation (1) is known as linear impulse or impulse of the force .Thus the final momentum impulse may be obtained by velocity adding initial momentum impulse during the time t impact as shown in fig. This is principle of impulse of movement. A collision between two bodies which occurs in very small interval of time during which the body exert large force of impact. The common normal to the surface contact during impact is known as line of impact of their mass center are located on this line the impact is said to direct central impact. If ball of mass m collide with concrete floor with velocity v, if ball is dropped with velocity then the centre of infinite mass floor also can be assume to be laying on the same line, hence the impact is direct central one nothing. The deformation phase that follow at the end of which the ball of floor will have regained their original shape of ball rebound back with velocity V , force F of external by floor on the ball are only impulsive force during the impact. The nature of variation are not instantly known in the externally small known interval of deformation of the restitution, hence these forces on only be average interval. The ratio of the impulse can be period of restitution to that can be the period of deformation is called coefficient denoted by e. Numerically e lies between zero of an depending on two material involved in the impact of the impulsive momentum can be written as e = ∫ R dt ∫ Pdt = mv ’ mv = v ’ v If a ball is dropped from height ‘H’ and it rebound back to a height ‘h’ V = √2gh , v’ = √2gh’ Hence, e = √h/H ………….(2) Procedure: 1. The height of the ball from floor before dropping is noted as H. 2. The ball also dropped of rebound is carefully noted h. 3. Step (1) and step (2) are separated there, for positive same H + average h is reached. 4. Steps (1), (2), & (3) are repeated for different height. Observation table: Sr. no Object H(m) h(m) e = √h/H Average 1 Rubber ball 2 Plastic ball Experiment no:3 Title: To find coefficient of restitution, impulse of rate of energy loss for a direct central impact between the two bodies. Div:__________ Roll No: ________ Date: ___________ Figure:-