Beore we start the new material we will do another Newton s second law problem. A bloc is being pulled by a rope as shown in the picture. The coeicient o static riction is 0.7 and the coeicient o inetic riction is 0.5. The bloc s mass is.0 g. The tension is 5 N. What happens? N T mg Solution: Does the bloc moe and i it does moe, what is its acceleration? The ree body diagram is aboe. The rictional orce opposes the motion o the bloc. I the tension is to the right, the bloc will also moe towards the right (i it moes). Applying Newton s second law (the -ais is horizontal and the y-ais is ertical) F T ma ma F y ma N mg ma y y Since the bloc can only moe along the surace, ay = 0 and the y-component gies N mg 0 N mg.0g9.8m/s 19.6 N Using the deinition or the coeicients o riction, the inetic riction will be N And the largest possible static riction will be, 0.5 19.6N 9.8 N 0.7 19.6N 13.7 N s, ma sn Since the tension (5 N) is greater than the largest possible static riction, the bloc will slide to the right. I the tension was less than s,ma then the bloc will not moe and the static riction would eactly equal the tension to preent the bloc rom moing. So i T = 10 N, s = 10 N. To ind the acceleration o the bloc Lesson 6, page 1
T a ma T m 5N 9.8N 7.6 m/s.0g Apparent Weight When you ride in a roller coaster, sometimes you eel heaier than normal, sometimes lighter. For an eleator accelerating up, you eel heaier, Fig. 04.55 But do you actually weigh more? No. Your weight does not change. What changes is the normal orce pushing up on you. Remember Newton s third law. The normal orce is the loor pushing up on you. The reaction is you pushing down on the loor. I the normal orce increases, so will your pushing down on the loor. What happens when the eleator accelerates downward? Now you eel lighter than normal. There is a problem with the two diagrams aboe. The ector representing the weight should hae the same length. They do not. Air Resistance is too complicated or us to wor with quantitatiely. Typically we will ignore the eects o air resistance. Lesson 6, page
Air resistance increases dramatically with speed. The aster an object goes, the greater the drag. Fundamental Forces Graity only attractie. Long range orce. Very wea. Electromagnetism eplained in olume Strong holds nucleus together Wea seen in some radioactie decay processes Chapter 5 Circular Motion In a rigid body, the distances between the parts o the body remain constant. We begin inestigating the rotation o a rigid body. We conclude our inestigation in Chapter 8. The language used to describe rotational motion is ery similar to the language used to describe linear motion. The symbols are dierent. Description Linear Angular position displacement rate o change o position, a a aerage rate o change o position t t lim lim instantaneous rate o change o position t 0 t t 0 t While we are amiliar with angles measured in degrees, we measure rotations in radians. In the igure, the angle measure in radians is deined as the ratio o the arc length s to the radius r: s r Lesson 6, page 3
For a complete rotation, s = r and the angle or a complete rotation is radians. This gies the conersion o radians to degrees 360 rad 180 rad The change in rotational position is eplained below. There is a relation between the speed o a point on the rim and the angular elocity o the wheel. s r s r s r t t r Just lie displacements, rotations hae directions. We tae counterclocwise rotations as positie and clocwise rotations as negatie. Other useul parameters The period (T) The time it taes or a compete reolution. The requency () The number o reolutions per unit time. Lesson 6, page 4
These parameters are related. By deinition 1 T A complete reolution is the circumerence r. The speed is the distance diided by the time. For a complete reolution, the time is the period T r T r Using the equation near the top o the page, we hae r r Rolling motion I the wheel rolls across the ground, its translational speed depends on how ast it spins. I the wheel rolls without slipping, the ale o the wheel moes with a speed gien by r To show this we note that this is a relatie elocity problem! A Ale o wheel, G Ground, R Rim o wheel. The rim is motionless on the ground when the wheel rolls without slipping. Lesson 6, page 5
AG AG AG r Radial Acceleration A particle undergoing uniorm circular motion ( is constant in time), still accelerates een though its speed is constant. The elocity continuously changes direction. AR AR AR RA 0 RG The change in elocity points towards the center o the circle. To ind the magnitude o the radial acceleration, consider the diagram below. The magnitude o length is ery close to the arc length. As shrins, it becomes equal to the arc arc length radius o circle angle subtended t Using the deinition o acceleration we hae a t t t Using = r rom aboe, we hae the two epressions or the radial acceleration r a r r Lesson 6, page 6
You need to memorize the epressions or the radial acceleration. The acceleration is perpendicular to the elocity as long as the magnitude o the elocity is constant (the speed is constant). Problem-Soling Strategy or an Object in Uniorm Circular Motion (page 163) 1. Begin as or any Newton s second law problem: identiy all the orces acting on the object and draw a FBD.. Choose perpendicular aes at the point o interest so that one is radial and the other is tangent to the circular path. 3. Find the radial component o each orce. 4. Apply Newton s second law as ollows: Fr ma r where Fr is the radial component o the net orce and the radial component o the acceleration is r a r r (For uniorm circular motion, neither the net orce nor the acceleration has a tangential component since the speed is constant.) Lesson 6, page 7