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Example: The simple pendulum

Example: The simple pendulum. Suppose we release a mass m from rest a distance h 1 above its lowest possible point. What is the maximum speed of the mass and where does this happen? To what height h 2 does it rise on the other side?. m. h 1. h 2. v. Example: The simple pendulum.

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Example: The simple pendulum

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  1. Example: The simple pendulum • Suppose we release a mass m from rest a distance h1 above its lowest possible point. • What is the maximum speed of the mass and wheredoes this happen? • To what height h2 does it rise on the other side? m h1 h2 v

  2. Example: The simple pendulum • Kinetic+potential energy is conserved since gravity is a conservative force (E = K + U is constant) • Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice)E = 1/2mv2 + mgy y h1 h2 y = 0 v

  3. Example: The simple pendulum • E = 1/2mv2 + mgy. • Initially, y = h1and v = 0, so E = mgh1. • Since E = mgh1initially, E = mgh1always since energyisconserved. y y = 0

  4. Example: The simple pendulum • 1/2mv2will be maximum at the bottom of the swing. • So at y = 0 1/2mv2 = mgh1v2 = 2gh1 y y = h1 h1 y = 0 v

  5. Example: The simple pendulum • Since E = mgh1 =1/2mv2 + mgy it is clear that the maximum height on the other side will be at y = h1 = h2and v = 0. • The ball returns to its original height. y y = h1 = h2 y = 0

  6. Example: The simple pendulum • The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of energy between K and U. E = 1/2mv2 + mgy= K + U = constant. y

  7. Generalized Work/Energy Theorem: • The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. Emechanical =K+U of system not conserved! • If all the forces are conservative, we know that K+U energy is conserved: K + U = Emechanical = 0 which says that WNC = 0. • If some non-conservative force (like friction) does work, K+Uenergy will not be conserved and WNC= E. WNC = K + U = Emechanical

  8. Problem: Block Sliding with Friction • A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is k. • How far, x, does the block go along the bottom portion of the track before stopping? d  k x

  9. Problem: Block Sliding with Friction... • Using WNC = K + U • As before, U = -mgd • WNC = work done by friction = -kmgx. • K = 0 since the block starts out and ends up at rest. • WNC = U -kmgx= -mgd x = d / k d k x

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