General Physics I. Lecture 9: Vector Cross Product. Prof. WAN, Xin ( 万歆 )

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1 General Physics I Lecture 9: Vector Cross Product Prof. WAN, Xin ( 万歆 ) xinwan@zju.edu.cn

2 Outline Examples of the rotation of a rigid object about a fixed axis Force/torque point of view Energy point of view Vector cross product Necessity of the new math Definition and examples Coriolis effect

Example: Atwood's Machine Two blocks having masses m1 and m 2 are connected to each other by a light cord that passes over two identical, frictionless pulleys, each

3 Example: Atwood's Machine Two blocks having masses m1 and m 2 are connected to each other by a light cord that passes over two identical, frictionless pulleys, each having a moment of inertia I and radius R. Find the acceleration of each block and the tensions T 1, T 2, and T 3 in the cord. (Assume no slipping between cord and pulleys.)

4 Example: Atwood's Machine

5 Example: Atwood's Machine Discussion: Equal mass: At equilibrium. Unequal mass: Normally we assume a direction for the acceleration. If the result is negative, the real acceleration is in the opposite direction. I = 0: Goes back to the same old Newton's laws.

6 Example: Connected Cylinders Consider two cylinders having masses m 1 and m 2, where m 1 ¹ m 2, connected by a string passing over a pulley. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest. Find the linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the pulley at this time.

7 Detailed Analysis We are now able to account for the effect of a massive pulley. Because the string does not slip, the pulley rotates. We neglect friction in the axle about which the pulley rotates for the following reason: Because the axle s radius is small relative to that of the pulley, the frictional torque is much smaller than the torque applied by the two cylinders, provided that their masses are quite different. Mechanical energy is constant; hence, the increase in the system s kinetic energy (the system being the two cylinders, the pulley, and the Earth) equals the decrease in its potential energy.

8 Example: Connected Cylinders

9 Brainstorming I How to understand t = rf sinf? Force is a vector. Position is a vector. Torque is also a vector.

10 Brainstorming II How to understand v = rw? Remember righthand rule? Velocity is a vector. Position is a vector. Angular velocity is also a vector.

11 Angular vs Linear Speed when a rigid object rotates about a fixed axis, every particle of the object moves in a circle whose center is the axis of rotation. v = ω r

12 The Vector Product The direction of C is perpendicular to the plane formed by A and B, and the best way to determine this direction is to use the right-hand rule.

13 Properties of Vector Product Unlike the scalar product, the vector product is not commutative. Instead, the order in which the two vectors are multiplied in a cross product is important: The vector product obeys the distributive law:

14 Orthogonal Unit Vectors If A is parallel or antiparallel to B, then A B = 0; therefore, it follows that A A = 0. If A is perpendicular to B, then A B = AB

15 Generic Results A= A x î+ A y ĵ+ A z k B=B x î+b y ĵ+b z k A x î B y ĵ= A x B y k A y ĵ B x î=a y B x k C= A B=C x î+c y ĵ+c z k =î A y A z z ĵ B y B + A z A x k = x B z B + A x A y î y B x B ĵ k A x A y A z B x B y B z

16 Derivative The derivative of the cross product with respect to some variable (such as t) is where it is important to preserve the multiplicative order of A and B

17 Acceleration: Vector Version For rotation of a rigid object about a fixed axis v = d r a = d v = ω r = d ( ω r ) = d ω r + ω d r The vector formulas have no requirement that r is perpendicular to the axis of rotation. = α r + ω ( ω r )

18 Angular vs Linear Acceleration when a rigid object rotates about a fixed axis, every particle of the object moves in a circle whose center is the axis of rotation.

19 Rotational Kinetic Energy The total rotational kinetic energy of the rotating rigid object is the sum of the kinetic energies of the individual particles

20 Rotational Kinetic Energy K = 1 2 i m i ω r i 2 = 1 2 i m i( ω r i ) ( ω r i ) r = r( r ω) ω ω = 1 2 i m i [ω 2 r i 2 ω 2 ( ω r i ) 2 ] r = 1 2 ( 2 i m i r i, )ω 2 = 1 2 I ω2 Identity: ( a b) ( c d)=( a c)( b d)( a d)( b c)

21 Typhoon (Low Pressure) Rotation: Clockwise in the southern hemisphere and counterclockwise in the northern hemisphere.

22 In a Rotating System Newton's Second Law: T=m v2 r Fictitious centrifugal force Tm v2 r =0

23 Coriolis Effect The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. v = ω r for point object at rest in the rotational frame What if it is moving?

24 Graphic Illustration In the inertia frame of reference d r I = d r R + ω r r + d r I ω r d r R r change of the displacement within the rotational frame change of the displacement due to the rotation of the reference frame

25 Mathematical Origin This is generically true for any vector u. d u I = d u R + ω u That is, we need to take into account the change of directions of the axes. d(u x î) = du x d î + u î x = du x x: î + ω (u x î) y: z: d(u y ĵ) d(u z k) = du y = du z ĵ + u y d ĵ k + u z d k = du y = du z ĵ + ω (u y ĵ) k + ω (u z k)

26 Acceleration as the 2nd Derivative In the inertia frame of reference v I = d r I d 2 r 2 I d v I = [ = d r R = d v R d R = d2 r 2 R + ω r + ω v + ω ]2 r + 2 ω d r R Δ r R r + Δ r I Δ r R r ω r ω r + ω ( ω r )

27 Force in the Rotating Frame In the rotating frame of reference, total force F net = m d2 r 2 R = m d2 r 2 I Coriolis force 2m ω d r R = F 2m ω v R m ω ( ω r ) centrifugal force m ω ( ω r ) = F 2m ω v R + mω 2 r for r ω Identity: a ( b c) = b( a c) c( a b)

28 Manifestation of Coriolis This low-pressure system (in Northern hemisphere) spins counterclockwise due to balance between the Coriolis force and the pressure gradient force.

29 Warning The treatment of the Coriolis force is to study the motion of an object in the rotating frame of reference (typically on another rigid object, like the earth, which rotates with a constant angular velocity). This is different from studying the rotation of a rigid object in the inertia frame of reference, which is the main subject of our interest. Once again, stay in the inertial frame of reference unless you cannot.

General Physics I. Lecture 9: Vector Cross Product. Prof. WAN, Xin ( 万歆 )

General Physics I. Lecture 9: Vector Cross Product. Prof. WAN, Xin ( 万歆 ) General Physics I Lecture 9: Vector Cross Product Prof. WAN, Xin ( 万歆 ) xinwan@zju.edu.cn http://zimp.zju.edu.cn/~xinwan/ Outline Examples of the rotation of a rigid object about a fixed axis Force/torque