Practice Exam 2 Solutions

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department o Physis Physis 801T Fall Term 004 Problem 1: stati equilibrium Pratie Exam Solutions You are able to hold out your arm in an outstrethed horizontal position thans to the ation o the deltoid musle Assume the humerus bone has a mass o m, the enter o mass o the humerus is a distane d rom the sapula, the deltoid musle attahes to the humerus a distane a distane s rom the sapula and the angle the deltoid musle maes with the horizontal is α A shemati representation o this ation loos as ollows: a) What is the tension T in the deltoid musle? b) What are the vertial and horizontal omponents o the ore exerted by the sapula (shoulder blade) on the humerus? Solution: Choose the unit vetors î to point horizontally to the right and ĵ vertially upwards Let F x and F x be the horizontal and vertial omponents o the ore exerted by the sapula (shoulder blade) on the humerus? The ondition that the sum o the ores ating on the rigid body is zero, r r r r r F total = T + mg + F = 0, beomes î : F T x ĵ : F + T y os α = 0 sin α mg = 0 1

2 Choose the ontat point o the sapula (shoulder blade) on the humerus to ompute the torque Then the total torque about this point S is Then torque equilibrium beomes r r r τ total S = τ S, T + τ S,m r = 0 The torque about S due to the enter o mass o the humerus is r r τ ˆ ( ˆ ˆ Sm = r S,m m g = di mgj) = dmg, 1 The torque about S due to the deltoid musle is r r τ ˆ ( ˆ ˆ ˆ ST = r S T T = si T os αi + T sin α j) = st sin α,, So the total omponent o the torque in the ˆ diretion must vanish st sinα dmg = 0 We an solve the torque equation or the magnitude o the deltoid musular ore dmg T = s sin α The horizontal ore equation an now be solved or the horizontal omponent o the ore o the sapula on the humerus F = T osα = dmg os α x ssin α The vertial ore equation an now be solved or the vertial omponent o the ore o the sapula on the humerus, F = mg T sinα = mg dmg y = mg (1 d s s ) < 0 Notie that the vertial omponent points downward sine d s > 1

3 Problem : water buet A water buet attahed to a rope and spun in a vertial irle o radius r Suppose the buet has mass m and that the buet an be approximated as a point mass at the end o the rope The rope breas when the buet is moving vertially upwards The buet rises to a height h above the release point a) What was the veloity o the buet when it was released? \ b) What was the tension in the ord when it broe? Solution: The ore diagram on the buet is shown in the igure Choose the initial state, at time t 0, when the string has just broen Choose the inal state, at time t, when the buet has reahed it s maximum height Choose zero or potential energy at the point where the string breas 1 Initial Energy: The initial ineti energy is K 0 = m v 0 The initial potential energy is U 0 = 0 So the initial mehanial energy is E 0 = K 0 + U = 1 m v 0 0 3

4 Final Energy: The inal ineti energy is K = 0 The inal potential energy is U = mgh So the inal mehanial energy is E = K +U = mgh Non-onservative wor: There is no non-onservative wor Change in Mehanial Energy: The hange in mehanial energy is then 0 = W = E meh, n 0 = mgh 1 mv 0 This equation an be solved or the esape veloity, v 0, v 0 = gh Just beore the string broe the radial ore equation is Thereore Newton s Seond Law in the radial diretion, rˆ, is T = mv 0 r So the tension in the string is T = mgh r 4

5 Problem 3: spring and loop A mass m is pushed against a spring with spring onstant and held in plae with a ath The spring ompresses an unnown distane x When the ath is removed, the mass leaves the spring and slides along a ritionless irular loop o radius r When the mass reahes the top o the loop, the ore o the loop on the mass (the normal ore) is equal to twie the weight o the mass a) Using onservation o energy, ind the ineti energy at the top o the loop Express your answer as a untion o, m, x, g, and R b) Using Newton s seond law, derive the equation o motion or the mass when it is at the top o the loop ) How ar was the spring ompressed? Solution: Choose polar oordinates with origin at is at the equilibrium position the enter o the loop Let x denote the U spring ( x = 0) = 0 displaement o the spring rom the equilibrium position Choose the zero point or the gravitational potential energy U grav = 0 at the bottom o the loop Choose the zero point or the spring potential energy when the spring Initial Energy: Choose or the initial state, the instant beore the ath is released The initial ineti energy K 0 = 0 The initial potential energy is non-zero, U 0 = (1 ) x So the initial mehanial energy is E 0 = K 0 + U 0 = (1 ) x Final Energy: Choose or the inal state, the instant the mass is at the top o the loop The inal ineti energy K = 1 mv sine the mass is in motion The inal potential energy is non-zero, U = mg R 5

6 So the inal mehanial energy is E = K +U = mgr + 1 mv Non-onservative Wor: Sine we are assuming the tra is ritionless, there is no non- onservative wor Change in Mehanial Energy: The hange in mehanial energy is thereore zero, 0 = W n = E mehanial = E E 0 Thus mehanial energy is onserved, E = E 0, or So the ineti energy at the top o the loop is 1 mgr + mv = 1 x 1 mv = 1 x mgr b) The ore diagram is shown in the igure Thereore Newton s Seond Law in the radial diretion, rˆ, is mg N = mv R When the mass reahes the top o the loop, the ore o the loop on the objet (the normal ore) is equal to twie the weight o the mass, N = mg Thereore the Seond Law beomes 3mg = mv R We an rewrite this ondition in terms o the ineti energy as 3 1 mgr = mv ) Substituting the ore ondition into the energy ondition that we ound in part a) yields, 6

7 7 1 mgr = x Thus the displaement o the spring rom equilibrium is x = 7mgR 7

8 Problem 4: inlined plane, rition, spring An objet o mass m slides down a plane whih is inlined at an angle θ The objet starts out at rest a distane d rom an unstrethed spring that lies at the bottom o the plane The spring has a onstant a) Assume the inline plane is ritionless How ar will the spring ompress when the mass irst omes to rest? b) Now assume that the inline plane has a oeiient o ineti rition µ How ar will the spring ompress when the mass irst omes to rest? ) In ase b), how muh energy has been lost to heat? Solution: Let x denote the displaement o the spring rom the equilibrium position Choose the zero point or the gravitational potential energy U grav = 0 at the bottom o the inlined plane but at the loation o the end o the unstrethed spring Choose the zero point or the spring potential energy when the spring is at the equilibrium position, U spring ( x = 0) = 0 Initial Energy: Choose or the initial state, the instant the objet is released at the top o the inlined plane The initial ineti energy K 0 = 0 The initial potential energy is nonzero, = mgdsinθ So the initial mehanial energy is U 0 E 0 = K 0 + U = mgdsinθ 0 8

9 Final Energy: Choose or the inal state, when the objet irst omes to rest and the spring is ompressed a distane x at the bottom o the inlined plane The inal ineti energy K = 0 sine the mass is in motion The inal potential energy is non-zero, U = 1 x xmg sin θ Notie that the gravitational potential energy is negative beause the objet has dropped below the height o the zero point o gravitational potential energy So the inal mehanial energy is E = K +U = 1 x xmg sinθ Non-onservative Wor: Sine we are assuming the tra is ritionless, there is no non- onservative wor Change in Mehanial Energy: The hange in mehanial energy is thereore zero, 0 = W n = E mehanial = E E 0 Thus mehanial energy is onserved, E = E 0, or This is a quadrati equation dmg sinθ = 1 x xmg sinθ mg sin θ dmg sin θ x x = 0 with positive hoie o square root or the solution to insure a positive displaement o the spring rom equilibrium, mg sin θ + m g sin θ dmg sin θ 1 x = + 9

10 b) The eet o ineti rition is that there is now a non-zero non-onservative wor done on the objet that has moved a distane, d + x, given by W = (d + x ) = µ N ( d + x ) = µ mg os θ (d + n x ) Note the normal ore is ound by using Newton s Seond Law in the diretion perpendiular to the inlined plane, N mg osθ = 0 Change in Mehanial Energy: The hange in mehanial energy is thereore, W = E = E E, n mehanial 0 whih beomes 1 µ mg osθ (d + x ) = x xmg sin θ dmg sin θ This simpliies to 1 0 = x xmg (sin θ µ os θ ) dmg (sin θ µ os θ ) This is the idential equation as in part a) above with So the new displaement o the spring is sinθ sin θ µ osθ mg ( sin θ µ os θ ) m g ( sin θ µ os θ ) dmg ( sin θ µ os θ ) x = ) The heat lost is given by W = µ mg osθ (d + n x ), where x is given in part b) 10

11 Problem 5: ellipti orbit A satellite o mass m is in an elliptial orbit around a planet o mass m p whih is loated at one ous o the ellipse The satellite has a veloity v a at the distane r a when it is urthest rom the planet The distane o losest approah is r p a) What is the magnitude o the veloity v p o the satellite when it is losest to the planet? b) I the satellite were in a irular orbit o radius r = r p, is it s veloity v greater than, equal to, or less than the veloity v p o the original ellipti orbit? Justiy Solution: your answer Choose zero or the gravitational potential energy to be when the satellite and planet are separated by an ininite distane, U r = =0 When the satellite is at losest approah the energy is ( ) 1 Gm m E = K +U = mv 1 p p p 1 p r p When the satellite is at the urthest distane rom the planet, the energy is 1 Gm m E = K +U = mv 1 a a a 1 a r a Mehanial energy is onserved so 1 Gm m 1 Gm m mv 1 = mv 1 1 p r 1 a p ra Thereore the veloity o the satellite at losest approah is 11

12 v 1 1 = v Gm r a r p p a b) The veloity o a satellite undergoing uniorm irular motion an be ound rom the ore equation, So the veloity is Gm 1 m = mv 1 r r v = Gm r Suppose we give the satellite that is in a irular motion a small inrease in veloity in the tangential diretion, v new = v + v This implies that the energy inreases The satellite will no longer travel in a irular orbit sine or the same radius r, mv Gm m r r 1 new > 1 The satellite will move away rom the planet entering into an elliptial orbit So any veloity v p greater than v will orm an ellipti orbit 1