In-Class Problems 22-23: Mechanical Energy Solution

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deparent o Physics Physics 801 TEAL Fall Term 004 In-Class Problems -3: Mechanical Energy Solution Section Table and Group Number Names Hand in one solution per group We would like each group to apply the problem solving strategy with the our stages (see below) to answer the ollowing two problems I Understand get a conceptual grasp o the problem II Devise a Plan - set up a procedure to obtain the desired solution III Carry our your plan solve the problem! IV Look Back check your solution and method o solution 1

2 Problem : Escape Velocity and Mechanical Energy The asteroid Toro, discovered in 1964, has a radius o about R = 50 km and a mass o about m t = kg Let s assume that Toro is a perectly uniorm sphere What is the ape velocity or an object o mass m on the surace o Toro? Could a person reach this speed (on Earth) by running? Solution: The only potential energy in this problem is the gravitational potential energy We choose the zero point or the potential energy when the object and Toro are an ininite distance apart, U gravity (r 0 = ) 0 Then the potential energy when the object and Toro are an ininite distance r apart is given by Gm U gravity (r) = with U gravity (r 0 = ) 0 r The expression ape velocity reers to the minimum velocity necessary or the object to ape the gravitational interaction o the asteroid and move o to an ininite distance away I the object has a velocity less than the ape velocity, it will be unable to ape the gravitational orce and must return to Toro I it has a velocity greater than the ape velocity, it will have a non-zero kinetic energy at ininity So the condition or the ape velocity is that the object will have exactly zero kinetic energy at ininity We choose our initial state, at time t 0, when the object is at the surace o the planet with velocity equal to ape velocity We choose our inal state, at time t, to occur when the separation distance between the asteroid and the object is ininite Initial Energy: The initial kinetic energy is K 0 = 1 mv The initial potential energy is Gm U 0 = So the initial mechanical energy is R E 0 = K 0 +U 0 = 1 mv R t Gm m Final Energy: The inal kinetic energy is K = 0, since this is the condition that deines ape velocity The inal potential energy is zero, U = 0 since we chose the zero point or potential energy at ininity So the inal mechanical energy is E = K + U = 0 Non-conservative work: There is no non-conservative work

3 Change in Mechanical Energy: The change in mechanical energy 0 = W = E mech, nc is then 1 Gm 0 = mv R This equation can be solved or the ape velocity, v, ( N m kg )(0 10 kg) Gm v = t = = 73 m s 3 R (50 10 m) 1 Considering that Olympic sprinters typically reach velocities o 1 m s 1, this is an easy velocity to attain by running on Earth It may be harder on Toro to generate the acceleration necessary to reach this speed by pushing o the ground, since any slight upward orce will raise the center o mass and it will take substantially more time than on earth to come back down or another push o the ground Problem 3: Circular Motion and Conservation o Mechanical Energy An object o mass m is released rom rest at a height h above the surace o a table The object slides along the inside o the loop-the-loop track consisting o a ramp and a circular loop o radius R shown in the igure Assume that the track is rictionless When the object is at the top o the track it pushes against the track with a orce equal to three times it s weight What height was the object dropped rom? Solution: 3

4 We choose polar coordinates with origin at the center o the loop We choose the zero point or the potential energy U = 0 at the bottom o the loop Initial Energy: We choose or our initial state, the instant the mass is released The initial kinetic energy K 0 = 0 The initial potential energy is non-zero, U 0 = mgh So the initial mechanical energy is E 0 = K 0 +U = mgh 0 Final Energy: We choose or our inal state, the instant the mass is at the top o the loopthe-loop The inal kinetic energy K = 1 mv since the mass is in motionat rest The inal potential energy is non-zero, U = mg R So the inal mechanical energy is E = K +U = mgr + 1 mv Non-conservative Work: Since we are assuming the track is rictionless, there is no non- conservative work Change in Mechanical Energy: The change in mechanical energy is thereore zero, 0 = W nc = E mechanical = E E 0 Thus mechanical energy is conserved, E = E 0, or mgr + 1 mv = mgh Missing Condition: The normal orce o the track on the object is perpendicular to the direction o the motion o the object so this orce does zero work, r N d r = 0 4

5 Thereore the work-kinetic energy theorem does not account or the action o this orce When there are orces that do no work in some direction, set up the Second Law in that direction, r r F = ma We show the orce diagram when the mass is at the top o the loop Thereore Newton s Second Law in the radial direction, rˆ, is mg N = m v R Notice rom the inormation given in the example, the normal orce o the loop on the r r object is N = 3mg (This is the action-reaction pair) Thereore the Second Law becomes 4mg = m v R We can rewrite this condition in terms o the kinetic energy as Summary: Our two equations are thereore mgr = 1 mv mgr + 1 mv = mgh 1 mgr = mv The second equation (rom the Newton s Second Law) can be substituted into the conservation o mechanical energy equation to yield, So the initial height is 4mgR = mgh h= 4R 5