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. . Perusing QUIZ. 1.If a molecule moves along a bend with a steady speed, then its tangential part of increasing speed isA)positive.B) negative.C)zero.D)constant.. 2.The typical segment of quickening representsA)the time rate of progress in the greatness of the velocity.B) the time rate of alter in the course of the velocity.C)magnitude of the velocity.D)direction of the aggregate acceleratio

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CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS (Section 12.7) Today\'s Objectives : Students will have the capacity to decide the typical and distracting segments of speed and increasing speed of a molecule going along a bended way. In-Class Activities : • Check homework, if any • Reading test • Applications • Normal and digressive segments of speed and increasing speed • Special instances of movement • Concept test • Group critical thinking • Attention test

READING QUIZ 1. If a molecule moves along a bend with a steady speed, then its distracting part of quickening is A) positive. B) negative. C) zero. D) constant. 2. The ordinary part of speeding up speaks to A) the time rate of progress in the greatness of the speed. B) the time rate of alter in the course of the speed. C) magnitude of the speed. D) direction of the aggregate speeding up.

APPLICATIONS Cars going along a clover-leaf trade encounter an increasing speed because of an adjustment in speed and additionally because of an adjustment in course of the speed. On the off chance that the auto\'s speed is expanding at a referred to rate as it goes along a bend, how might we decide the size and bearing of its aggregate increasing speed? Why might you think about the aggregate increasing speed of the auto?

APPLICATIONS (proceeded with) A bike goes up a slope for which the way can be approximated by a capacity y = f(x). On the off chance that the cruiser begins from rest and expands its speed at a steady rate, how might we decide its speed and increasing speed at the highest point of the slope? How might you dissect the bike\'s "flight" at the highest point of the slope?

NORMAL AND TANGENTIAL COMPONENTS When a molecule moves along a bended way, it is once in a while helpful to depict its movement utilizing organizes other than Cartesian. At the point when the way of movement is known, typical (n) and unrelated (t) directions are frequently utilized. In the n-t organize framework, the inception is situated on the molecule (the beginning moves with the molecule ). The t-hub is digression to the way (bend) at the moment considered, positive toward the molecule\'s movement. The n-pivot is opposite to the t-hub with the positive heading toward the focal point of bend of the bend.

NORMAL AND TANGENTIAL COMPONENTS (proceeded with) The positive n and t headings are characterized by the unit vectors u n and u t , separately. The focal point of ebb and flow , O\', dependably lies on the inward side of the bend. The span of arch , r , is characterized as the opposite separation from the bend to the focal point of ebb and flow by then. The position of the molecule at any moment is characterized by the separation, s, along the bend from a settled reference point.

The size is dictated by taking the time subsidiary of the way work , s(t). v = v u t where v = s = ds/dt . Speed IN THE n-t COORDINATE SYSTEM The speed vector is dependably digression to the way of movement (t-bearing). Here v characterizes the size of the (speed) and u t characterizes the course of the speed vector.

Acceleration is the time rate of progress of speed : a = d v/dt = d(v u t )/dt = v u t + v u t . . . Here v speaks to the adjustment in the size of speed and u t speaks to the rate of alter in the course of u t . . After scientific control, the speeding up vector can be communicated as: . a = v u t + (v 2/r ) u n = a t u t + a n u n . Speeding up IN THE n-t COORDINATE SYSTEM

• The unrelated segment is digression to the bend and toward expanding or diminishing speed. a t = v or a t ds = v dv . Quickening IN THE n-t COORDINATE SYSTEM (proceeded) There are two parts to the speeding up vector: a = a t u t + a n u n • The typical or centripetal segment is constantly coordinated toward the focal point of ebb and flow of the bend. a n = v 2/r • The greatness of the increasing speed vector is a = [(a t ) 2 + (a n ) 2 ] 0 . 5

1) The molecule moves along a straight line . r => a n = v 2/r = 0 => a = a t = v . The extraneous segment speaks to the time rate of progress in the extent of the speed . 2) The molecule moves along a bend at consistent speed. a t = v = 0 => a = a n = v 2/r . The ordinary segment speaks to the time rate of alter in the course of the speed. Exceptional CASES OF MOTION There are some uncommon instances of movement to consider.

4) The molecule moves along a way communicated as y = f(x). The range of shape, r, anytime on the way can be computed from r = ________________ [ 1 + (dy/dx) 2 ] 3/2 d 2 y/dx 2 SPECIAL CASES OF MOTION (proceeded with) 3) The digressive part of speeding up is consistent , a t = (a t ) c . For this situation, s = s o + v o t + (1/2)(a t ) c t 2 v = v o + (a t ) c t v 2 = (v o ) 2 + 2(a t ) c (s – s o ) As some time recently, s o and v o are the underlying position and speed of the molecule at t = 0. How are these conditions identified with shot movement conditions? Why?

EXAMPLE PROBLEM Given: Starting from rest, a speedboat goes around a round way of r = 50 m at a speed that increments with time, v = (0 . 2 t 2 ) m/s. Find: The sizes of the watercraft\'s speed and quickening at the moment t = 3 s. Arrange: The vessel begins from rest (v = 0 when t = 0). 1) Calculate the speed at t = 3s utilizing v(t). 2) Calculate the unrelated and ordinary segments of increasing speed and after that the extent of the quickening vector.

. 2) The speeding up vector is a = a t u t + a n u n = v u t + (v 2/r ) u n . . Extraneous segment : a t = v = d( . 2t 2 )/dt = 0 . 4t m/s 2 At t = 3s: a t = 0 . 4t = 0 . 4(3) = 1 . 2 m/s 2 EXAMPLE (proceeded with) Solution: 1) The speed vector is v = v u t , where the size is given by v = (0 . 2t 2 ) m/s. At t = 3s: v = 0 . 2t 2 = 0 . 2(3) 2 = 1 . 8 m/s Normal part : a n = v 2/r = (0 . 2t 2 ) 2/( r) m/s 2 At t = 3s: a n = [ (0 . 2)(3 2 )] 2/(50 ) = 0 . 0648 m/s 2 The size of the quickening is a = [(a t ) 2 + (a n ) 2 ] 0 . 5 = [(1 . 2) 2 + (0 . 0648) 2 ] 0 . 5 = 1 . 20 m/s 2

CONCEPT QUIZ 1. A molecule going in a roundabout way of span 300 m has an immediate speed of 30 m/s and its speed is expanding at a consistent rate of 4 m/s 2 . What is the greatness of its aggregate speeding up at right now? A) 3 m/s 2 B) 4 m/s 2 C) 5 m/s 2 D) -5 m/s 2 2. If a molecule moving in a roundabout way of span 5 m has a speed work v = 4t 2 m/s, what is the size of its aggregate quickening at t = 1 s? A) 8 m/s B) 8.6 m/s C) 3.2 m/s D) 11.2 m/s

GROUP PROBLEM SOLVING Given: A fly plane goes along a vertical explanatory way characterized by the condition y = 0 . 4x 2 . At point A, the fly has a speed of 200 m/s, which is expanding at the rate of 0 . 8 m/s 2 . Find: The extent of the plane\'s increasing speed when it is at point A. Arrange: 1) The change in the speed of the plane (0.8 m/s 2) is the distracting segment of the aggregate increasing speed. 2) Calculate the sweep of ebb and flow of the way at A. 3) Calculate the typical segment of speeding up. 4) Determine the greatness of the increasing speed vector.

1) The digressive part of quickening is the rate of increment of the plane\'s speed, so a t = v = 0 . 8 m/s 2 . . dy/dx = d(0 . 4x 2 )/dx = 0 . 8x, d 2 y/dx 2 = d (0 . 8x)/dx = 0 . 8 At x =5 km, dy/dx = 0 . 8(5) = 4, d 2 y/dx 2 = 0 . 8 => r = ________________ = [ 1 + (4) 2 ] 3/2/(0 . 8 ) = 87 . 62 km [ 1 + (dy/dx) 2 ] 3/2 d 2 y/dx 2 GROUP PROBLEM SOLVING (proceeded with) Solution: 2) Determine the range of shape at point A (x = 5 km): 3) The ordinary part of quickening is a n = v 2/r = (200) 2/(87 . 62 x 10 3 ) = 0 . 457 m/s 2 4) The greatness of the quickening vector is a = [(a t ) 2 + (a n ) 2 ] 0 . 5 = [(0 . 8) 2 + (0 . 457) 2 ] 0 . 5 = 0 . 921 m/s 2

ATTENTION QUIZ 1. The extent of the ordinary speeding up is A) proportional to span of shape. B) inversely relative to span of ebb and flow. C) sometimes negative. D) zero when speed is consistent. 2. The bearings of the unrelated increasing speed and speed are dependably A) perpendicular to each other. B) collinear. C) in the same direction. D) in inverse bearings.

CURVILINEAR MOTION: CYLINDRICAL COMPONENTS (Section 12.8) Today\'s Objectives: Students will have the capacity to decide speed and increasing speed parts utilizing round and hollow directions. In-Class Activities: Check homework, if any Reading test Applications Velocity Components Acceleration Components Concept test Group critical thinking Attention test

1. In a polar arrange framework, the speed vector can be composed as v = v r u r + v θ u θ = r u r + r q u q . The term q is called A) transverse velocity. B) spiral speed. C) rakish velocity. D) precise quickening. . . . 2. The speed of a molecule in a round and hollow facilitate framework is A) r B) r q C) ( r q) 2 + ( r) 2 D) ( r q) 2 + ( r) 2 + ( z) 2 . . . . . . . Perusing QUIZ

In the figure appeared, the kid slides down the slide at a steady speed of 2 m/s. How quick is his height from the beginning (i.e., what is z )? . APPLICATIONS The round and hollow organize framework is utilized as a part of situations where the molecule moves along a 3-D bend.

APPLICATIONS (proceeded with) A polar organize framework is a 2-D representation of the tube shaped arrange framework. At the point when the molecule moves in a plane (2-D), and the spiral separation, r, is not steady, the polar arrange framework can be utilized to express the way of movement