Physics 1501: Lecture 9

1. Physics 1501: Lecture 9 • Announcements • Homework #3 : due next Monday • But HW 04 will be due the following Friday (same week). • Midterm 1: Monday Oct. 3 • Topics • Review Friction • Circular motion and Newton’s Laws

2. m2 T1 m1 m3 Example Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. What is the magnitude and direction of acceleration on the three blocks ? What is the tension on the two cords ?

3. N=-m2g T12 T23 T12 m2 a T23 T1 m2 mk m2g a a m1 m2g m1g m3g m3 m1 m3 T12 T23 T12 T23 T23 - m3g = m3a T12 - m1g = - m1a -T12 + T23 + mk m2g = - m2a SOLUTION: T12 = = 30.0 N , T23 = 24.2 N , a = 2.31 m/s2 left for m2

4. Lecture 9, ACT 1Friction and Motion • A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 30 N. The box 1 is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=3.5 and mk= 0.5. This second box can slide on an ice rink (frictionless). • The acceleration of box 1 is • Greater than (b) Equal to (c) Smaller than the acceleration of box 2 ? a1 friction coefficients ms=3.5 andmk=0.5 T m1 a2 m2 slides without friction

5. T m1 f m1g N1  T - f = 0 T = f  No sliding  T=Ma a1=a2=T/(m1+m2) ANSWER: (b) Lecture 9, ACT 1Solution • First, we need to know if box 1 will slide on box 2 • FBD of box 1 • Box 1 does not slide if • The maximum value for the static friction force is • f MAX = msN1 = msm1g • Box 1 does not slide if f = T ≤ f MAX • Here f MAX = msm1g = 3.5  1 kg  10 m/s2 = 35 N f = T = 30 N ≤ f MAX = 35 N Both boxes have the same acceleration

6. T m1 f m1g N1 m1 T f N1 -T -f m1g -N1 -m1g Lecture 9, ACT 1More details • Let us find how FBD and Newton’s laws lead to our intuitive result • FBD of box 1 • Action/reaction pairs (Newton’s 3rd law)

7. -N1 -f N1 f -f m2 -N1 m2 N2 m2g N2 m2g -N2 -m2g Lecture 9, ACT 1More details • FBD of box 2 • Action/reaction pairs (Newton’s 3rd law)

8. a1 T m1 f T- f= m1 a1  -N1 -f N1 = m1 g m1g N1 m2 m2g f = m2 a2 N2  j N2 = (m1 +m2)g a1= a2 = a (no sliding) i Lecture 9, ACT 1More details • Box 1 i : T + f = m1a1 j : N1 + m1g = 0 • Box 2 i : -f = m2a2 j : N2 - N1 + m2g = 0 a2 T= f+m1 a1= m2a2+m1 a1= (m2+m1)a a= 30 N / (1 kg + 2 kg) = 10 m/s2

9. Lecture 9, ACT 2Friction and Motion • Consider the same boxes as ACT 1, but now the box 1 is being pulled by a horizontal string having tension T = 40 N. Now, the acceleration of box 1 is • Greater than (b) Equal to (c) Smaller than the acceleration of box 2 ? a1 friction coefficients ms=3.5 andmk=0.5 T m1 a2 m2 slides without friction

10. T m1 f m1g N1  T - f = 0 T = f  It will slide ANSWER: (b) Lecture 9, ACT 2Solution • First, we need to know if box 1 will slide on box 2 • FBD of box 1 • Box 1 does not slide if • The maximum value for the static friction force is • f MAX = msN1 = msm1g • Box 1 does not slide if f = T ≤ f MAX • Here f MAX = msm1g = 3.5  1 kg  10 m/s2 = 35 N T = 40 N > f MAX = 35 N Both boxes have different accelerations • The maximum value for the static friction force is • ANSWER: (a)

11. T m1 f m1g N1 Lecture 9, ACT 2Solution • The force of friction is therefore due to the kinetic coefficient of friction k • The FBD and force pairs are the same as before, except that we now have for f f = mk| N1 | = mk m1 g • So only the i component will be modified.

12. a1 T m1 f -N1 -f m1g N1 m2 m2g N2 j  T + f = m1a1 T- f= m1 a1  a1=(T- mk m1 g) /m1  a1 =(40 N- 0.5  1kg 10 m/s2) /1kg = 35 m/s2 i -f = m2a2  f = m2 a2  a2 = mk m1 g / m2  a2 = 0.5  1kg 10 m/s2 /2 kg = 2.5 m/s2 Lecture 9, ACT 2More details • Box 1 • Box 2 a2

13. aC Newton’s Laws and Circular Motion v • Centripedal Acceleration • aC = v2/R • What is Centripedal Force ? • FC = maC = mv2/R R Animation

14. Example Problem I am feeling very energized while I shower. So I swing a soap on a rope around in a horizontal circle over my head. Eventually the soap on a rope breaks, the soap scatters about the shower and I slip and fall after stepping on the soap. To decide whether to sue ACME SOAP I think about how fast I was swinging the soap (frequency) and if the rope should have survived. From the manufacturers web site I find a few details such as the mass of the soap is 0.1 kg (before use), the length of the rope is 0.1 m and the rope will break with a force of 40 N. (assume FBS is large versus the weight of the soap)

15. Example Problem Step 1 In this case the picture was given. I need to find the frequency of the soap’s motion that caused the rope to break. I will use Newton’s Second Law and uniform circular motion.

16. Example Problem Step 2 Diagram. v T I will solve for the frequecy f. { I know M=0.1 kg, R=0.1m, and FBS = 40N } I will use F = ma (Newton’s Second Law), a=v2/r=w2r (circular motion) 2pf = w

17. Example Problem Step 3 – Solve Symbolically 1. SF = T = ma 2. a = w2R  T = m w2R 3. w = 2pf  T = 4p2f2mR In general When it breaks

18. Example Problem Step 4 – Numbers

19. Example Problem • Step 5 – Analyze • The units worked out to give s-1, which is correct for a frequency. • It seems that the suit is in trouble, if not necessarily dead. Being able to twirl your soap safely 10 rev/s is pretty good. I might have been excessively boisterous.

20. Lecture 9, ACT 3Circular Motion Forces • How fast can the race car go ? (How fast can it round a corner with this radius of curvature ?) mcar = 1500 kg mS = 0.5 for tire/road R = 80 m A) 10 m/s B) 20 m/s C) 75 m/s D) 750 m/s R

21. Banked Corners In the previous ACT, we drew the following free body diagram for a race car going around a curve on a flat track. N Ff mg What differs on a banked curve ?

22. N ma Ff mg Banked Corners Free Body Diagram for a banked curve. For small banking angles, you can assume that Ff is parallel to ma. This is equivalent to the small angle approximation sinq = tanq. Can you show that ?

23. Nonuniform Circular Motion Earlier we saw that for an object moving in a circle with nonuniform speed then a = ar + at . at ar What are Fr and Ft ?

24. Example Exercise 1 My match box car is going to do a loop the loop. What must be its minimum speed at the top so that it can make the loop successfully ??

25. q mg q N q q mg Example Exercise 1 Radial : Fr = N + mg cosq = mv2/R Tangential : Ft = mg sinq Solve the first for v.

26. N q mg Example Exercise 1 To stay on the track there must be a non-zero normal force. Why ? The limiting condition is where N = 0. And at the top, q = 0.