2008 Quarter-Final Exam Solutions

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1 2008 Quate-final Exam - Solutions Quate-Final Exam Solutions 1 A chaged paticle with chage q and mass m stats with an initial kinetic enegy K at the middle of a unifomly chaged spheical egion of total chage Q and adius R q and Q have opposite signs The spheically chaged egion is not fee to move (a) Find the value of K 0 such that the paticle will just each the bounday of the spheically chaged egion (b) How much time does it take fo the paticle to each the bounday of the egion if it stats with the kinetic enegy K 0 found in pat (a)? Solution: Assume that q is negative and Q is positive (a) Apply Gauss s Law to a spheical shell of adius whee < R Then, whee the chage density ρ = 3Q 4πR 3 E4π 2 = 4πρ3 ɛ 0 Solving fo the Electic Field at a distance fom the cente, we find E = ρ 3ɛ 0 We now find the potential diffeence between the cente of the sphee and the oute bounday of the chaged cloud V = R Substituting in ou expession fo E, we now find 0 Ed Using consevation of enegy, V = ρr2 6ɛ 0 K lost = U gained K 0 = q V = qq 8πRɛ 0 (b) Apply Newton s Second Law to the object of chage q when it is located a distance away fom the cente of the sphee

2 2008 Quate-final Exam - Solutions 2 F net = ma ( q)e = m d2 dt 2 d 2 dt 2 = qρ 3ɛ 0 m We ecognize that this is the diffeential equation fo simple hamonic motion whee ω = qρ 3ɛ 0 m d 2 dt 2 = ω2 Since the chage has the minimum kinetic enegy needed to each the suface, the tip fom the cente of the sphee to the oute bounday is one-fouth of a cycle of SHM Theefoe, Substituting, in fo ho, we find t = T 4 = π 2ω = π 3ɛ0 m 2 qρ t = π 2 4πɛ 0 mr 3 2 A unifom pool ball of adius begins at est on a pool table The ball is given a hoizontal impulse J of fixed magnitude at a distance β above its cente, whee 1 β 1 The coefficient of kinetic fiction between the ball and the pool table is µ You may assume the ball and the table ae pefectly igid Ignoe effects due to defomation (The moment of inetia about the cente of mass of a solid sphee of mass m and adius is I cm = 2 5 m2 ) qq (a) Find an expession fo the final speed of the ball as a function of J, m, and β (b) Fo what value of β does the ball immediately begin to oll without slipping, egadless of the value of µ Solution 1: (a) Conside an axis pependicula to the initial impulse and coplana with the table (Thoughout this solution we conside only toques and angula momenta with espect to this axis) Afte the initial impulse, the toque about this axis is always zeo, so angula momentum is conseved The initial impulse occus a pependicula distance (β + 1) fom the axis, so the angula momentum is L = (β + 1) J Afte the ball has skidded along the table fo a cetain distance, it will begin to oll without slipping At that point, ω = v f

3 2008 Quate-final Exam - Solutions 3 Meanwhile, its moment of inetia about this axis is (by the paallel-axis theoem) I = I cm + m 2 = 7 5 m2, so that its final angula velocity ω is given by L = Iω (β + 1) J = 7 5 m2 ω Theefoe, (β + 1) J = 7 5 mv f v f = 5J (1 + β) 7m (b) The ball acquies linea momentum J as a esult of the hoizontal impulse, so its initial velocity v is given by mv = J We want the initial angula velocity and initial velocity to satisfy the no-slip condition v = ω; thus (β + 1) J = 7 5 J Solution 2: β = 2 5 Conside toques and angula momenta about the cente of mass If the hoizontal impulse is lage compaed with the hoizontal impulse fom fiction duing the time that the cue stick is in contact with the ball, then angula impulse = change in angula momentum becomes: Jβ = 2 5 m2 ω 0 Linea impulse = change in linea momentum yields: J = mv 0 (b) We want the initial angula velocity and initial velocity to satisfy the no-slip condition v 0 = ω 0 ; thus Jβ = 2 5 mv 0 = 2 5 J β = 2 5

4 2008 Quate-final Exam - Solutions 4 (a) In the case that the ball does not immediately begin to oll without slipping, fiction will exet an angula impulse, ft, about the cente of mass of the ball as the ball skids along the suface: ft = 2 5 m2 (ω f ω 0 ) Fiction will also exet a linea impulse ft that will cause a change in linea momentum: Combining the last two equations: ft = m(v f v 0 ) m(v f v 0 ) = 2 5 m(ω f ω 0 ) Afte the ball has skidded along the table fo a cetain distance, it will begin to oll without slipping At that point, ω = v f v f + v 0 = 2 5 v f 2 5 ω 0 Now, using the elationships between the impulse J and the initial angula and linea momentum, v f + J m = 2 5 v f Jβ m J m (1 + β) = 7 5 v f v f = 5J (1 + β) 7m 3 A block of mass m slides on a cicula tack of adius whose wall and floo both have coefficient of kinetic fiction µ with the block The floo lies in a hoizontal plane and the wall is vetical The block is in constant contact with both the wall and the floo The block has initial speed v 0 (a) Let the block have kinetic enegy E afte taveling though an angle θ Deive an expession fo de dθ in tems of g,, µ, m and E (b) Suppose the block cicles the tack exactly once befoe coming to a halt Detemine v 0 in tems of g,, and µ

5 2008 Quate-final Exam - Solutions 5 Solution Let v be the speed of the block afte it has taveled though an angle θ When we apply Newton s Second Law to the block as it is taveling aound the cicula path of adius at a speed v, we find that the foce of the wall on the block, F w, is and the foce of the floo on the block, F N, is F w = mv2 F N = mg Theefoe, the total foce of fiction on the block is F F ic = µm( v2 + g) Let de denote the loss of kinetic enegy as a esult of the wok W that fiction does on the block when it has taveled though an angle dθ The block will tavel a distance ds = dθ as it tavels though an angle dθ Then, de = W = F F ic ds = F F ic dθ Now, substituting in the expessions fo F F ic, we find de = µm(v 2 + g)dθ de dθ = µ(mv2 + mg) = µ(2e + mg) whee we have simply used the fact that the kinetic enegy E is E = mv2 2 (b) Now, we sepaate vaiables and solve the diffeential equation Exponentiating both sides, yields de 2E + mg = µdθ 1 ln 2E + mg = µθ + C 2 2E + mg = Ce 2µθ Let E 0 denote the initial kinetic enegy Then, using initial conditions, we find that C = 2E 0 + mg Now, using the fact that E = 0 when θ = 2π, we obtain mg = (2E 0 + mg)e 4πµ

6 2008 Quate-final Exam - Solutions 6 Solving fo E 0, Then, using E 0 = mv2 0 2, and we find that E 0 = 1 2 mg(e4µπ 1) v 0 = g(e 4µπ 1) 4 Two beads, each of mass m, ae fee to slide on a igid, vetical hoop of mass m h The beads ae theaded on the hoop so that they cannot fall off of the hoop They ae eleased with negligible velocity at the top of the hoop and slide down to the bottom in opposite diections What is the maximum value of the atio m/m h such that the hoop always emains in contact with the gound? Neglect fiction Solution 1: Daw a fee-body diagam fo each bead; let F N be the (inwad) nomal foce exeted by the hoop on the bead Let θ be the angula position of the bead, measued fom the top of the hoop, and let the hoop have adius We see that F N + mg cos θ = m v2 F N = m v2 mg cos θ The (downwad) vetical component F Ny is given by F Ny = F N cos θ Fom Newton s thid law, the two beads togethe exet an upwads vetical foce on the hoop given by F u = 2F Ny ( ) v 2 F u = 2m cos θ g cos θ noting that the beads clealy each the same position at the same time Meanwhile, when each bead is at a position θ it has moved though a vetical distance (1 cos θ) Thus fom enegy consevation, 1 2 mv2 = mg (1 cos θ) Inseting this into the pevious esult, v 2 = 2g (1 cos θ) F u = 2m cos θ (2g (1 cos θ) g cos θ) F u = 2mg ( 2 cos θ 3 cos 2 θ )

7 2008 Quate-final Exam - Solutions 7 If the beads eve exet an upwad foce on the hoop geate than m h g, the hoop will leave the gound; ie, the condition fo the hoop to emain in contact with the gound is that fo all θ, F u m h g We can eplace the left side by its maximum value Letting s = cos θ, The deivative is zeo at s = 1 3, whee F u = 2mg ( 2s 3s 2) d ds F u = 2mg (2 6s) F u(max) = 2 3 mg So ou condition is 2 3 mg m hg m 3 m h 2 Solution 2: As befoe, we apply enegy consevation to find the speed of the beads: v = 2g (1 cos θ) The vetical (downwad) component of the beads velocity is thus v y = v sin θ v y = sin θ 2g (1 cos θ) While the hoop is in contact with the gound, the beads ae the only pat of the system in motion, so the momentum of the system is simply the beads momentum: The net (downwad) foce on the system is p y = 2mv y p y = 2m sin θ 2g (1 cos θ) F net = dp y dt = dp y dθ dθ dt = dp y dθ ) F net = 2m 2g d ( sin θ 1 cos θ 1 2g (1 cos θ) dt ( F net = 4mg cos θ ) cos θ + sin θ 2 1 cos θ sin θ cos θ v ( F net = 4mg cos θ cos 2 θ sin2 θ F net = 2mg ( 2 cos θ 3 cos 2 θ + 1 ) )

8 2008 Quate-final Exam - Solutions 8 This downwad foce is povided by gavity and the nomal foce upwad on the hoop: F net = 2mg + m h g F N If the hoop is to emain in contact with the gound, the nomal foce can neve be negative: F N 0 2mg + m h g F net 0 2mg + m h g 2mg ( 2 cos θ 3 cos 2 θ + 1 ) 0 2mg ( 2 cos θ 3 cos 2 θ ) m h g and we poceed as above

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