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 Example: Coriolis effect Torque

Example: Rotating Rod A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical

3 Example: Rotating Rod A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end?

4 Example: Rotating Rod The torque due to this force about an axis through the pivot is The moment of inertia for this axis of rotation Can you show?

5 Example: Rotating Rod The linear acceleration of the right end of the rod This result that at > g for the free end of the rod is rather interesting. It means that if we place a coin at the tip of the rod, hold the rod in the horizontal position, and then release the rod, the tip of the rod falls faster than the coin does!

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

6 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.)

7 Example: Atwood's Machine

8 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.

9 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.

10 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.

11 Example: Connected Cylinders

Torque Again The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque r: the distance between the pivot point

12 Torque Again The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque r: the distance between the pivot point and the point of application of F. d: the perpendicular distance from the pivot point to the line of action of F. This quantity d is called the moment arm of F.

13 Vector Representation The torque t involves the two vectors r and F, and its direction is perpendicular to the plane of r and F. We can establish a mathematical relationship between t, r, and F, using a new mathematical operation called the vector product, or cross product Will discuss more next week.

14 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.

15 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:

16 More Properties 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 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 Elementary Results

18 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 k A x B x A y y B =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

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

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

21 Coriolis Effect The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. Vector form for point object at rest in the rotational frame

22 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

23 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)

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

25 Force in the Rotating Frame In the rotating frame of reference F net = m d2 r 2 R = m d2 r 2 I Coriolis force 2m ω d r R centrifugal force m ω ( ω r ) = F 2m ω v R 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)

26 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 if you can.

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