1. PHY 231 1 PHYSICS 231 Lecture 29: Problems Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom

2. PHY 231 2 Figure skating A male figure-skater rotates around a vertical axis (1 rev/s) while holding his partner at 1.5 m. He then pulls her towards him (1 m distance). A) How fast will he rotate now? Assume no net torque is exerted on the skaters. B) If the woman is 55 kg, how much work has the man done while pulling? A) Use conservation of angular momentum: L i =I i  i =M partner r i 2  i = M*1.5 2 *2  *1=14.1M kgm 2 /s L f =I f  f =M partner r f 2  f =M*1 2 *  f L i =L f so,  f =14.1 rad/s=2.25 rev/s B) Consider the change in rotational KE initial: KE=½I  i 2 =½mr i 2  i 2 =0.5*55*1.5 2 *(2  ) 2 =2443 J final: KE=½I  f 2 =½mr f 2  f 2 =0.5*55*1 2 *14.1 2 =5467 J difference: 3024 J: equals the work done by the man.

3. PHY 231 3 Ball on a upward slope A ball (r=0.1 m) is rolled up a slope. Its linear velocity at the base is 10 m/s. How high will it go? (assume the ball does not slip and given I ball =2/5*Mr 2. Use conservation of mechanical energy. Initial: ½mv 2 +½I  2 +mgh= ½mv 2 +½2/5mr 2 (v/r) 2 +mgh= =0.7mv 2 =70m Final: ½mv 2 +½I  2 +mgh=mgh=9.8mh KE final =KE initial so h=70/9.8=7.1 m

4. PHY 231 4 Ball game In a children’s game at a fair, buckets are placed at 1m from the center around which they all rotate (I total =50kg/m 2, 1rev/s). The children try to throw balls in the buckets. After 50 balls have gone in the buckets, the wheel has slowed down to 0.5 rev/s. How heavy is one ball? 1 rev/s Top view Use conservation of angular momentum: Initial L=I  =50*2  =100  Final L=I  =(50+N balls m balls r)*  =(50+50*m*1)  L initial =L final so 50m=50 m=1 kg.

5. PHY 231 5 Bulk modulus A rubber ball, with bulk modulus B, volume V at 1 atm. increases its volume by 1 cm 3 when put in a vacuum chamber (P=0). If a ball of the same material but 5 times larger in volume at 1 atm, is put under a pressure of 3 atm, how much will its volume shrink? B=-  P/(  V/V) First case (1 atm -> vacuum): B=(-1 atm)/(-1 cm 3 /V) Second case (1 atm -> 3 atm): B=(2 atm)/(  V/5V) B is a constant, so must be equal in both cases:  V=10 cm 3 If you are not sure whether you need to convert to SI units, just do it: it is a bit of extra work, but at least you are sure it’s okay.

6. PHY 231 6 Buoyant forces When submerged in water an object weighs 1.6N. At the same time, the water level in the water container (with A=0.01 m 2 ) rises 0.01 m. What is the specific gravity (sg) of the object? (  water =1.0x10 3 kg/m 3 ) A=0.01 m 2 Use the fact that the Buoyant force on a submerged object equals the weight of the displaced water. W=F g -B =M object g-M water,displaced g =  object V object g-  water V object g =V object g (  object -  water ) 1.6N=0.01*0.01*g(  object -  water )=1.0x10 -4 *9.8*  water (sg-1) sg=2.63

7. PHY 231 7 Keep it coming. A plastic bag contains a glucose solution. The part of the bag that is not filled is under vacuum. If the pressure in a blood vein is 1.33x10 4 Pa, how high must one hang the bag to make sure the solution (specific gravity 1.02) enters the body? (  w =1.0x10 3 kg/m 3 ) P=P 0 +  gh 1.33x10 4 =0+1.02*1.0x10 3 *9.8*h h=1.33 m

8. PHY 231 8 Titanic: After the Titanic sunk, 10 people manage to seek refuge on a 2x4m wooden raft. It is still 1.0 cm above water. A heavy debate follows when another person (60 kg) wants to board as well. Fortunately, a PHY231 student is among the 10. Can she convince the others that itis safe to pull the person on board without the whole raft sinking? (  w =1.0x10 3 kg/m 3 ) With 10 people: F g =B (M raft +M 10 )g=V displaced,before  w g With 11 people: F g =(M raft +M 10 +M 1 )g B=(V displaced,before +V extra )  w g stationary if F g =B (M raft +M 10 )g+M 1 g=(V displaced,before +V extra )  w g M 1 g=V extra  w g so V extra =(M 1 /  w ) V extra =60/1.0x10 3 =0.06m 3 V extra =LxWxH=8H=0.06 so H=0.75cm so there is still 0.25cm to spare!

9. PHY 231 9 Bernoulli A=5cm 2 A=2cm 2, P=1 atm 2m Water flows over a height of 2m through an oddly shaped pipe. A) If the fluid velocity is 1 m/s at the bottom, what is it at the top? B) What is the water pressure at the top? A) Use the equation of continuity: A 1 v 1 =A 2 v 2 5*v top =2*1 v top =0.4 m/s B) Use Bernoulli. P top +½  v top 2 +  gh top = P bot +½  v bot 2 +  gh bot P top +0.5*(1E+03)*0.4 2 +(1E+3)*9.8*2=(1E+05)+0.5*(1E+03)1 2 P top =80820 Pa.  =1.0x10 3 kg/m 3

10. PHY 231 10 Centipede A centipede (100 legs) decides to cross a lake. The contact length between each of his legs and the water is 1mm. If its mass is 2g and  water =0.073 N/m will he succeed? Assume that the angle  =45 0. FsFs FsFs FgFg  1 leg F g =0.002*9.8=0.0196N F surface tension,vertical =100*2  water L contact sin  =200*0.073*0.001*0.707 =0.010N F g >F surface tension, vertical : he sinks!

11. PHY 231 11 Ch. 10 Metal hoop A metal (thermal expansion coefficient  =17x10 -6 1/ 0 C) hoop of radius 0.10 m is heated from 20 0 C to 100 0 C. By how much does its radius change? 0.1m  L=  L 0  T =17x10 -6 (2  r 0 )80=8.5x10 -4 m r new =(L 0 +  L)/2  =L 0 /2  +  L/2  =r 0 +1.35x10 -4 m

12. PHY 231 12 10: Diving Bell A cylindrical diving bell (diameter 3m and 4m tall, with an open bottom is submerged to a depth of 220m in the sea. The surface temperature is 25 0 C and at 220m, T=5 0 C. The density of sea water is 1025 kg/m 3. How high does the sea water rise in the bell when it is submerged? Consider the air in the bell. P surf =1.0x10 5 Pa V surf =  r 2 h=28.3m 3 T surf =25+273=298K P sub =P 0 +  w g*depth=2.3x10 6 Pa V sub =? T sub =5+273=278K Next, use PV/T=constant P surf V surf /T surf =P sub V sub /T sub plug in the numbers and find: V sub =1.15m 3 (this is the amount of volume taken by the air left) V taken by water =28.3-1.15=27.15m 3 =  r 2 h h=27.15/  r 2 =3.8m rise of water level in bell.

13. PHY 231 13 10: Moles Two moles of Nitrogen gas (N 2 ) are enclosed in a cylinder with a moveable piston. A) If the temperature is 298 K and the pressure is 1.01x10 6 Pa, what is the volume (R=8.31 J/molK)? b) What is the average kinetic energy of the molecules? k B =1.38x10 -23 J/K A)PV=nRT V=nRT/P =2*8.31*298/1.01x10 6 =4.9E-03 m 3 B) E kin,average =½mv 2 =3/2k B T=3/2*1.38x10 -23 *298=6.2x10 -21 J

14. PHY 231 14 Ch 11: 3*10 3 J of heat is transferred to a 1cm 3 cube of gold at T=20 o C Will all the gold have melted afterwards? L f =6.44x10 4 J/kg T melt =1063 o C c specific =129 J/kg 0 C  =19.3x10 3 kg/m 3. When Solid: m=  V=(19.3x10 3 )x(1x10 -6 m 3 )=1.93x10 -2 kg. Q=cm  T=129*1.93x10 -2 *  T=2.5  T To raise to melting point:  T=(1063-20)=1043 o C, so Q=2608J During the phase change from solid to liquid: Q=L f m Q=6.44x10 4 *1.93x10 -2 =1.24x10 3 J to make liquid. Total needed: 3850 J, only 3000 J available, doesn’t melt completely.

15. PHY 231 15 11: Heat transfer 3 A hot block and a cold block are thermally connected. Three different methods to transfer heat are proposed as shown. Which one is the most efficient way (fastest) to transfer heat fro hot to cold and what are the relative rates of transfer? Area: A Length L 0.1A 0.2L 4A 5L A: cross section surface of black wire,L:its length Use: P=kA  T/L Case 1: P~A/L Case 2: P~0.1A/0.2L=0.5A/L Case 3: P~4A/5L=0.8L P 1 :P 2 :P 3 = 1:0.5:0.8 First case is most efficient. 1 2

16. PHY 231 16 Thermal equilibrium 20g of a solid at 70 0 C is placed in 100g of a fluid at 20 o C. After waiting a while the temperature of the whole system is 30 o C and stays that way. The specific heat of the solid is: a)Equal to that of the fluid b)Less than that of the fluid c)Larger than that of the fluid d)Unknown; different phases cannot be compared e)Unknown; different materials cannot be compared Q fluid =-Q solid m fluid c fluid (T final -T fluid )=-m solid c solid (T final -T solid ) C fluid -m solid (T final -T solid ) C solid m fluid (T final -T fluid ) == -20(30-70) 100(30-20) = 0.8 C solid >C fluid