Physics 170 Week 9 Lecture 2

Physics 170 Week 9 Lecture 2 http://www.phas.ubc.ca/ gordonws/170 Physics 170 Week 9 Lecture 2 1

Textbook Chapter 1: Section 1.6 Physics 170 Week 9 Lecture 2 2

Learning Goals: We will solve an example problem using tangential-normal coordinates. We will formulate Newton s second law in a cylindrical coordinate system. We shall study an example where Newton s law in cylindrical coordinates is used to solve a problem. You should also understand how to relate the orientations of the tangential-normal coordinate system and the cylindrical coordinate system for the case where the motion is entirely in a single plane. We will solve an example, where one needs to know how to do this. Physics 170 Week 9 Lecture 2

Example: The smooth block B, having a mass of 0.2 kg, is attached to the vertex A of the right circular cone using a light cord. The cone is rotating at a constant angular rate about the z axis such that the block attains a speed of 0.5 m/s. At this speed, determine the tension in the cord and the reaction which the cone exerts on the block. Neglect the size of the block. Physics 170 Week 9 Lecture 2 4

Free Body Diagram Physics 170 Week 9 Lecture 2 5

Find a Mathematical description of the force vectors Gravity: W = mg û b Normal reaction force N = N ) 4 5 ûn + 5 ûb Tension in cord: T = T ) 5 ûn + 4 5 ûb Physics 170 Week 9 Lecture 2 6

Find a Mathematical description of the acceleration vector The acceleration vector is at) = vt) û t + v2 ρ ûn and we know that v = 0, ρ = 5 200 mm) and v = 0.5 m/s. a = 0.5 m/s)2 ûn 5 0.200 m) Physics 170 Week 9 Lecture 2 7

Impose Newton s Second Law where F = W + N + T = mg û b +N F = m a 4 5 ûn + ) 5 ûb +T Remember that we computed the acceleration and found a = Putting this together, we have 0.2 kg) +N 0.5 m/s)2 ûn 5 0.200 m) 0.5 m/s)2 5 0.200 m) ûn = 0.2 kg)9.81 m/s 2 ) û b 4 5 ûn + 5 ûb ) + T 5 ûn + 4 5 ûb 5 ûn + 4 ) 5 ûb ) Physics 170 Week 9 Lecture 2 8

Impose Newton s Second Law cont d We arrived at the formula 0.2 kg) +N 0.5 m/s)2 5 0.200 m) ûn = 0.2 kg)9.81 m/s 2 ) û b ) 4 5 ûn + 5 ûb + T 5 ûn + 4 ) 5 ûb First, consider the û b components. This gives an equation 5 N + 4 5 T = 0.2)9.81) N Then, consider the û n components. They give the formula 4 5 N + 5 T = 0.2 kg) 0.5)2 5 0.200) N Physics 170 Week 9 Lecture 2 9

Impose Newton s Second Law cont d Imposing Newton s law has resulted in the two equations 5 N + 4 5 T = 0.2)9.81) N 4 5 N + 5 Or, in matrix form [ ] [ /5 4/5 N 4/5 /5 T T = 0.2 kg) 0.5)2 5 0.200) N ] = [ 0.2)9.81) 0.2)0.5) 2 5 0.200) Which we solve by multiplying by the inverse matrix to get [ ] [ ] [ ] N /5 4/5 0.2)9.81) = 0.2)0.5) T 4/5 /5 2 N 5 0.200) ] N Physics 170 Week 9 Lecture 2 10

We have arrived at the equation [ ] [ N /5 4/5 = T 4/5 /5 Doing matrix multiplication gives ] [ 0.2)9.81) 0.2)0.5) 2 5 0.200) ] N N = 5 0.2)9.81) 4 5 T = 4 5 0.2)9.81) + 5 0.2)0.5) 2 5 0.200) N 0.2)0.5) 2 5 0.200) N N = 0.844 N, T = 1.82 N Physics 170 Week 9 Lecture 2 11

Review of cylindrical coordinates Position: rt) = rt) û r + zt) ˆk Velocity: vt) = ṙt) û t + rt) θt) û θ + żt) ˆk Acceleration: at) = rt) rt) θt) ) 2 û r + 2ṙt) θt) ) + rt) θt) û θ + zt) ˆk Physics 170 Week 9 Lecture 2 12

The Cartesian cylindrical dictionary If you know the Cartesian vector r = x î + y ĵ + z ˆk, we can identify the components of the same vector in the cylindrical coordinate system using r = x 2 + y 2, θ = arctan y x, z = z Alternatively, if we know the cylindrical coordinates of a point r, θ, z) we can identify the Cartesian coordinates x = r cos θ, y = r sin θ, z = z Also, the unit vectors are related by û r = cos θ î + sin θ ĵ, û θ = sin θ î + cos θ ĵ Physics 170 Week 9 Lecture 2 1

Equation of Motion in Cylindrical Coordinates The equation of motion F = m a is an equation for vectors. When the vectors are written in cylindrical coordinates F = F r û r + F θ û θ + F z ˆk, a = ar û r + a θ û θ + a z ˆk the equation of motion is an equation for each of the three components F r = ma r, F θ = ma θ, F z = ma z If there is more than one force acting on the particle, F is the resultant, that is, the vector sum of the forces. Physics 170 Week 9 Lecture 2 14

Example: The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, θ = 0.7t) rad, and z = 0.5t) m, where t is in seconds. Determine the components of force F r, F θ, and F z which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy. Physics 170 Week 9 Lecture 2 15

Strategy for solving: We are given the position as time-dependent cylindrical coordinates. From the position, we will find the velocity and the acceleration by taking time derivatives. Once we know the acceleration, we will use Newton s second law, F = m a in cylindrical coordinates to find the components of F. Physics 170 Week 9 Lecture 2 16

Solution: We are given the time dependent cylindrical coordinates: rt) = 1.5 m, θt) = 0.7)t rad, zt) = 0.50)t m where t is in seconds. From these, we find ṙt) = 0, θt) = 0.7) rad, żt) = 0.50) m rt) = 0, θt) = 0, zt) = 0 Physics 170 Week 9 Lecture 2 17

Solution cont d: We have found r, θ, z) = 1.5 m, 0.7)t rad, 0.50)t m), ṙ, θ, ż) = 0, 0.7) rad/s, 0.50) m/s), r, θ, z) = 0, 0, 0). We want to plug these into at) = rt) rt) θt) 2) û r + 2ṙt) θt) ) + rt) θt) û θ + zt) ˆk = 1.5 m)0.7 rad/s) 2 û r Physics 170 Week 9 Lecture 2 18

Solution cont d: We have found that the boy has acceleration. at) = 1.5 m)0.7 rad/s) 2 û r Apply Newton s second law F = m a to get the net force on the boy F = 40 kg)1.5 m)0.7 rad/s) 2 û r = 29.4 û r N Physics 170 Week 9 Lecture 2 19

Solution cont d: We have found that the net force on the boy is F = 40 kg)1.5 m)0.7 rad/s) 2 û r = 29.4 û r N This must be the resultant of the reaction of the slide F s and gravity W = mg ˆk: F = Fs mg ˆk. We find F s = 29.4 û r + 40 kg)9.81 m/s 2 ) ˆk N and F sr = 29.4 N, F sθ = 0, F sz = 92 N. Physics 170 Week 9 Lecture 2 20

Example: The 0.5 lb particle is guided along the circular path using the slotted arm guide. If the arm has an angular velocity θ = 4 rad/s and an angular acceleration θ = 8 rad/s 2 at the instant when θ = 0 degrees, determine the force of the guide on the particle. Motion occurs in the horizontal plane. Physics 170 Week 9 Lecture 2 21

Strategy for solving: We are given the angular speed and acceleration. We need to find rt) in terms of θt) in order to find r, ṙ and r at the instant in question. We will find rt) by first using geometry to find fθ). Then, we will use the result and information about θ, θ and θ to find r, ṙ and r. Physics 170 Week 9 Lecture 2 22

Strategy for solving cont d: We then use above information to write down the acceleration in cylindrical coordinates. Newton s second law tells us that the sum of all forces acting on the particle are equal to the mass times the acceleration. We can therefore multiply the acceleration that we have found by the mass to get the total force acting on the particle. Physics 170 Week 9 Lecture 2 2

Strategy for solving cont d: The reaction force of the guide is normal to the trajectory, and is easiest to describe in tangential-normal coordinates where forces are F = N û n + F û θ with F the force of the guide. Since motion occurs in the horizontal plane, we neglect gravity. We then equate the reaction and friction force to the total force that we found using Newton s second law. Physics 170 Week 9 Lecture 2 24

Strategy for solving cont d: In order to proceed, We shall have to find a relationship between the tangential-normal unit vectors û t and û n and the cylindrical unit vectors û r and û θ in order to solve this problem. Once we have the relation between unit vectors, we can solve for the normal reaction force in terms of known quantities. Physics 170 Week 9 Lecture 2 25

Solution: Let us find rθ). From the diagram we see that rθ) = 1.0 ft) cos θ Physics 170 Week 9 Lecture 2 26

Solution cont d: Using rθ) = 1.0 ft) cos θ, ṙθ) = 1.0 ft) θ sin θ ) rθ) = 1.0 ft) θ sin θ + θ2 cos θ) At the instant in question, θ = π/6 rad, θ = 4 rad/s, θ = 8 rad/s 2, r = 1.0ft) 2 = 2 ft, ṙ = 1.0)4)1 ft/s = 2 ft/s 2 r = 1.0) 8 2 + 42 ) = 2 4 + 8 ) ft/s 2 Physics 170 Week 9 Lecture 2 27

Solution cont d: The acceleration in cylindrical coordinates is at) = rt) rt) θt) 2) û r + 2ṙt) θt) ) + rt) θt) where we have put the z-component to zero. Plugging in: at) = 4 8 ) 2 42 ) û r + 2 2)4) + ) 2 8) û θ û θ ft/s 2 Physics 170 Week 9 Lecture 2 28

Solution cont d: The acceleration is at) = 4 8 at) = ) 2 42 ) 4 16 ) û r + û r + 2 2)4) + 16 + 4 ) ) 2 8) û θ ft/s 2 û θ ft/s 2 Physics 170 Week 9 Lecture 2 29

Solution cont d: Newton s second law is F = m a Inserting the mass 0.5 2.2 previous page, Nû n + F û θ =.5 2.2 slug) and the acceleration from the 4 16 ) û r +.5 2.2 16 + 4 ) û θ lb To calculate N, we take the dot-product of both sides with û r. We need to know û r û n. Physics 170 Week 9 Lecture 2 0

Theoretical Interlude When the motion is confined to a plane, that is, when the xy-plane, equivalently rθ-plane, is the osculating plane, it might be useful to know the relative orientations of the tangential-normal and the spherical polar coordinates: To find it, we write the velocity in tangential-normal and cylindrical coordinates v = v û t, v = ṙ û r + r θ û θ These are the same vector. The dot product with û t is û t û r = cos ψ = ṙ ṙ 2 + r 2 θ, tan ψ = r θ 2 ṙ = r dθ dr Physics 170 Week 9 Lecture 2 1

Theoretical Interlude cont d Summary: For motion in the xy=rθ-plane: [ ] [ ût û r û t û θ cos ψ sin ψ = û n û r û n û θ sin ψ cos ψ ] cos ψ = ṙ ṙ 2 + r 2 θ 2, sin ψ = tan ψ = r dθ dr = r dr/dθ r θ ṙ 2 + r 2 θ 2 Physics 170 Week 9 Lecture 2 2

Solution cont d: Nû n + F û θ =.5 4 16 ) û r +.5 2.2 2.2 We take the dot-product of both sides with û n to get ) Nû r û n =.5 2.2 N sin ψ) =.5 2.2 4 40 4 16 ) 16 + 4 ) û θ lb Physics 170 Week 9 Lecture 2

Solution cont d: We have found N sin ψ) =.5 2.2 4 16 ) Now tan ψ = r θ/ṙ =, sin ψ = 2, cos ψ = 1 2 ) 2 N = 0.569 lb N =.5 2.2 4 + 16 Physics 170 Week 9 Lecture 2 4

Solution cont d: Nû n + F û θ =.5 4 16 ) û r +.5 2.2 2.2 We take the dot-product of both sides with û t to get F û t û θ =.5 4 16 ) û t û r +.5 2.2 2.2 ) F =.5 2.2 4 16 cot ψ) +.5 2.2 16 + 4 ) 16 + 4 16 + 4 û θ lb ) û t û θ ) lb Physics 170 Week 9 Lecture 2 5

Solution cont d: F =.5 2.2 F =.5 2.2 F = 0.14 lb 4 16 ) cot ψ) +.5 2.2 4 16 ) 1 +.5 2.2 16 + 4 ) 16 + 4 ) lb lb Physics 170 Week 9 Lecture 2 6

For the next lecture, please read Textbook Chapter 14: Section 14.1-2 Physics 170 Week 9 Lecture 2 7