Solutions to Homework Set #5 Phys2414 Fall 2005


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1 Solution Set #5 1 Solutions to Homewok Set #5 Phys414 Fall 005 Note: The numbes in the boxes coespond to those that ae geneated by WebAssign. The numbes on you individual assignment will vay. Any calculated quantities that involve these vaiable numbes will be boxed as well. 1. GRR1 5.P.00. A socce ball of diamete 31 cm olls without slipping at a linea speed of.6 m/s. Though how many evolutions has the socce ball tuned as it moves a linea distance of 16 m? The distance the ball tavels, x, is equal to the cicumfeence of the ball, C, times the numbe of evolutions, α, it tuns. x = C α The cicumfeence of a sphee at its equato with diamete, d, is πd. The numbe of evolutions the ball tuned is then x = πd α α = x πd α = 16 m π 0.31 m α = 16.4 ev. GRR1 5.P.006. An elevato cable winds on a dum of adius 70.0 cm that is connected to a moto. (a) If the elevato is moving down at 0.60 m/s, what is the angula speed of the dum? The speed of the elevato must be the speed of the ope as it leaves the dum. Since the ope is unspooling fom the dum, this is also the speed of the dum at its edge. The equation which elates angula speed with linea speed is The angula speed of the dum is then v = ω v 0.60 m/s m 0.86 ad/s (b) If the elevato moves down 9.0 m, how many evolutions has the dum made?
2 Solution Set #5 We may use the elationship we found in poblem 1 to find the numbe of evoltutions made. x = πd α α = x πd α = 9.0 m π( m α =.0 ev 3. GRR1 5.P.011. The appaatus of the figue below is designed to study insects at an acceleation of magnitude 960 m/s (= 98 g). The appaatus consists of a.0 m od with insect containes at eithe end. The od otates about an axis pependicula to the od and at its cente. (a) How fast does an insect move when it expeiences a centipetal acceleation of 960 m/s? Using the elationship between centipetal acceleation, linea speed, and adius, we can find the linea speed of the insect as it swings aound in this device. a c = v v = a c v = v = 31.0 m/s 960 m/s ( 1.0 m ) (b) What is the angula speed of the insect? The angula speed can be calculated using the cousin of the elationship above. a c = ω ac 960 m/s 1/.0 m 31.0 ad/s
3 Solution Set # GRR1 5.P.01. The oto (the figue below) is an amusement pak ide whee people stand against the inside of a cylinde. Once the cylinde is spinning fast enough, the floo dops out. (a) What foce keeps the people fom falling out the bottom of the cylinde? The feebody diagam of a peson in the oto looks like the following figue. F f +y F N + W The foce that acts against the weight of the peson in the fiction foce on the peson by the wall. Since the peson is not sliding, we know it s static fiction that is holding the peson up. (b) If the coefficient of fiction is 0.4 and the cylinde has a adius of.3 m, what is the minimum angula speed of the cylinde so that the people don t fall out? (Nomally the opeato uns it consideably faste as a safety measue.) The sums of the foces in the y and diections ae ΣF y = F f W = 0
4 Solution Set #5 4 ΣF = F N = ma c whee a c is the centipetal acceleation. The minimum angula speed of the cylinde would occu when the maximum fictional foce is exactly equal to the weight of the peson. Using the equation fo static fiction, F f = µ s F N, we can combine the two above equations to get F f = W µ s F N = mg µ s (ma c ) = mg µ s a c = g Recognizing that the centipetal acceleation may be expessed as a c = ω, we can ewite the last equation as µ s ω = g ω = g µ s g µ s The minimum angula velocity is then 9.8 m/s m 3. ad/s 5. GRR1 5.P.018. A highway cuve has a adius of 11 m. At what angle should the oad be banked so that a ca taveling at 6.3 m/s has no tendency to skid sideways on the oad? [Hint: No tendency to skid means the fictional foce is zeo.] The feebody diagam fo this situation looks like the following figue.
5 Solution Set #5 5 F N +y + θ W Notice that one of the coodinate axes is pointed in the diection of the acceleation. The sums of the foces in the y and diections ae ΣF y = F N W = 0 ΣF = F N sin θ = ma c Solving fo F N in the fist equation gives F N = W F N = mg F N = mg We can substitute this expession fo F N into the second equation above. This gives F N sin θ = ma ( ) c mg sin θ = ma c mg sin θ = ma c sin θ = a c g The faction on the left is equal to tan θ. We can also expess a c in tems of the linea speed, v, and the adius,. We can now solve fo theta. tan θ = v g θ = actan ( ) v g
6 Solution Set #5 6 The angle at which the ca does not skid sideways is then θ = ( ) ( 6.3 m/s) actan 9.8 m/s 11 m θ = GRR1 5.P.00. A olle coaste ca of mass 330 kg (including passenges) tavels aound a hoizontal cuve of adius 37 m. Its speed is 14 m/s. What is the magnitude and diection of the total foce exeted on the ca by the tack? The feebody diagam fo this situation is vey simila to that of the last poblem. In this case we will conside the only two foces acting on the ca, i.e. the foce on the ca by the tack, F CT, and the foce of gavity on the ca, W. F CT +y + θ W The sums of the foces in the y and diections ae then ΣF y = F CT,y W = 0 ΣF = F CT, = ma c The magnitude of the foce on the ca by the tack is given by F CT = (F CT, ) + (F CT,y ) Using the fist two equations we know that F CT, and F CT,y ae equal to The magnitude is then F CT, = ma c, F CT,y = mg F CT = (ma c ) + (mg) F CT = m a c + g F ( ) v CT = m + g
7 Solution Set #5 7 The angle above hoizontal is then given by tan θ = F CT,y F CT,x tan θ = mg ma c tan θ = g a c tan θ = g v ( ) g θ = actan v The magnitude and diection ae then F ( ) CT = 330 kg ( 14 m/s) + (9.8 m/s ) 37 m F CT = 3700 N θ = actan θ = 61.6 ( 9.8 m/s ) 37 m ( 14 m/s) 7. GRR1 5.P.03. A ca dives aound a cuve with adius 40 m at a speed of 33 m/s. The oad is banked at 5.. The mass of the ca is 1600 kg. (a) What is the fictional foce on the ca? This poblem is simila to poblem 5, but this time thee is a fictional foce. The coesponding feebody diagam is given in the figue below. F N +y F f + θ W
8 Solution Set #5 8 The sums of the foces in the y and diections ae then ΣF y = F N F f sin θ W = 0 ΣF = F N sin θ + F f = ma c We would like to solve fo the fictional foce to get the answe. Compaing the two equations above we can see that both equations have F N as the fist tem, but one is multiplied by while the othe is multiplied by sin θ. In ode to solve this system of equations, we have to eliminate F N. To do this we can multiply the fist equation by sin θ and the second equation by. Doing this gives the following two equations F N sin θ F f sin θ = W sin θ F N sin θ + F f cos θ = ma c Next we can multiply the fist equation by 1 and add the equations togethe F N sin θ + F f sin θ = W sin θ F N sin θ + F f cos θ = ma c Adding the equations togethe gives F f cos θ + F f sin θ = ma c W sin θ Factoing out F f on the left side we find F f (cos θ + sin θ) = ma c W sin θ The quantity in paentheses is a tigonometic identity and is equal to one. Finally we have F f = ma c W sin θ F f = m v mg sin θ ( ) v F f = m g sin θ The foce of fiction is then equal to F f = ( ) ( 33 m/s) 1600 kg 40 m cos( 5. ) 9.8 m/s sin( 5. ) F f = 710 N
9 Solution Set #5 9 (b) At what speed could you dive aound this cuve so that the foce of fiction is zeo? This speed may be found by setting the fictional foce in ou above expession equal to zeo. ( ) v 0 = m g sin θ 0 = v g sin θ Solving fo v we get v = g sin θ v = g sin θ v = g tan θ The velocity at which no fictional foce occus is then v = (9.8 m/s )( 40 m) tan( 5. ) v = 19.6 m/s
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