A B Ax Bx Ay By Az Bz

Transcription

1 Lecture 5.1 Dynamics of Rotation For some time now we have been discussing the laws of classical dynamics. However, for the most part, we only talked about examples of translational motion. On the other hand, while studying kinematics we have also learned about rotation. Let us look at dynamics of this process. In order to do that, we will have to learn some additional properties of vectors. 1. Vector Multiplication By now we have already discussed various physical quantities and we know that some quantities can only have magnitudes and called scalars and other quantities may have both magnitude and direction and these are called vectors. We have also learned that the special steps have to be taken in order to add vectors. Both magnitude and direction have to be taken into account when adding, so vector addition can not be reduced to a simple algebraic addition. Today we shall talk about multiplying vectors. Once again vector multiplication is very much different from regular scalar multiplication. There are two ways of how one can multiply vectors. 1) By definition, the scalar product of two vectors A and B (also known as the dot product) is A B ABcos, (5.1.1) where A and B are the magnitudes of these vectors and is the angle between them. It is easy to see that this quantity is a scalar, which has the significance of the projection of vector A on vector B or the projection of vector B on vector A. It can be positive as well as negative depending on the sign of the cosine-function. It reaches its maximum value, when vectors A and B are parallel, which means that the angle between them is equal to zero and cosine of this angle equals to 1. The scalar product becomes zero in the case of perpendicular vectors. It is easy to see that commutative law works for the scalar product, which is A B B A. (5.1.) The scalar product can be expressed in terms of vectors' components as A B Ax Bx Ay By Az Bz. (5.1.3) Exercise: Prove equation

2 In addition to a scalar product of two vectors, one can introduce the mathematical object, known as a vector product of two vectors. We shall call vector C A B to be the vector product (or the cross product) of vectors A and B if it has the magnitude C ABsin, (5.1.4) where is the smaller of the two angles between vectors A and B. The direction of this vector C is defined according to the right-hand rule. We have already mentioned this rule when talking about right-handed coordinate system. According to the rule, you have to place vectors A and B tail to tail without altering their orientations and consider the line perpendicular to the plane formed by these vectors. Place your right hand around this line in such a way that your fingers would sweep A into B through the smaller angle between them. Then your outstretched thumb points the direction of C. It is easy to see from this rule that commutative law does not work for the vector product, instead the anti-commutative law takes place B A d A Bi. (5.1.5) Another important property of the vector multiplication is that if vectors A and B are parallel or anti-parallel, their vector product is equal to zero. The vector product has its maximum magnitude if vectors A and B are perpendicular. In particular, if one considers the unit vectors along the coordinate axes of the right-handed coordinate system, it gives k i, j i j k, j k i. (5.1.6) This means that the vector product of two vectors can be expressed in terms of the vector components as d y z y zi b z x z xg d x y x yi A B A B B A i A B B A j A B B A k. (5.1.7) Exercise: Prove equation 5.1.7, using equation

3 . Torque and Newton's second law for rotation Let us consider rotation of a body around the fixed axis of rotation. We have already seen that the most effective way to describe kinematics of this motion is by means of the angular variables, such as angular displacement, angular velocity and angular acceleration. Now we have to introduce a similar set of variables in order to be able to describe dynamics of this motion. The main subject of dynamics is force. Let us see what the angular analog of this quantity is. In the case of translational motion, it does not matter at which point of the body you apply the force as long as it is a rigid body which moves as a whole. Indeed if you consider a heavy box on the floor, you can either pull it from the front or push it from behind with the same force and the result is going to be the same. However, it is different in the case of rotation. For instance, you may notice that a doorknob is always located as far from the door's hinge as possible. If you push this door right near the hinge, you will never be able to open it. On the other hand even very heavy doors can be opened without difficulty if you push them for the doorknobs. So, in the case of rotation it matters not only what force you apply, but also how far from the rotational axis you apply it. Direction also matters. If you push or pull this door in the direction along its surface it will not do any good for opening of the door. You can open it only if your force has a component in the direction perpendicular to the door. In general case of a rigid body rotating around fixed rotational axis, some force F may be applied in arbitrary direction at the point, which is located at distance r from the body's rotational axis. The position of this point is defined by the radius vector, r, which connects the pivot point (rotational axis) with that point. Then we shall define the torque of the force F as the cross product of two vectors r F. (5.1.8) The magnitude of this torque can be found as rf sin, (5.1.9) where is the smaller angle between the vectors r and F. If vector r and force F form the plane perpendicular to the rotational axis, torque will be parallel or antiparallel to the rotational axis. In a same way one can define the torque for a particle in a circular motion. We said that force F is acting in the arbitrary direction, which means it may have a radial component F r along the vector r as well as the tangential component F the vector r. This means that F sin perpendicular to

4 where r rf r F, (5.1.10) r sin is the perpendicular distance between the rotational axis and an extended line running through the vector F. This extended line is called line of action and r is called the moment arm of the force F. In our example about the door, we saw that if a force acts perpendicular to the door, it will be most efficient way to open it. In this case torque of that force reaches its maximum magnitude. If the force acts along the door it does not move the door, but in that case the torque of this force is equal to zero. In a same way it is equal to zero if the force is applied directly to the hinge, because r=0 and the door does not move either. So torque plays the same role for the rotation as force does for translation. The SI unit of torque is Newton-meter (Nm). Since torque is a vector, it obeys the superposition principle. So if there are several forces acting on a body, one can find torques for each of these forces and calculate the net torque with respect to the chosen rotational axis. The body will rotate around this axis with non-zero angular acceleration if the net torque is not zero in same way as the body will move with non-zero acceleration only if the net force acting on it is not zero. On the other hand rotation with constant angular speed does not require a non-zero net torque as it was in the case of the Newton's first law for translation. However, it still requires a centripetal force, which has zero torque. Now it is time to obtain mathematical equation for the Newton's second law in the case of rotation. First, we will do so for the particle-like mass, m, rotating around the circle of radius r. Let F be the net force acting on this particle. This force can have two components. It for sure should have the normal component towards the center of the circle, which provides centripetal acceleration for this particle. It may also have tangential component F, which is responsible for the angular acceleration and tangential acceleration. Newton's second law for this component gives or multiplying by radius F ma t, (5.1.11) rf rmat rmbrg, I, (5.1.1) where I mr is the moment of inertia for a particle. Equation has exact same form as Newton's second law, if one replaces the net force with the net torque, the acceleration with angular acceleration and mass with moment of inertia. So, moment of inertia, I, is the angular analog of

5 mass. Equation has been derived for a particle, but it can be generalized to the case of the rigid body of arbitrary shape. Indeed, any body can be considered as combination of particles of different masses m i at different distances r i from the rotational axis. One can perform the summation of equations for these particles, which gives i mi ri, i i (5.1.13) I, where angular acceleration is the same for all of the particles in the rigid body, so it can be taken out of summation. The net torque is a vector sum of all the torques for individual particles and moment of inertia of the rigid body is I m r, (5.1.14) i i i where r i are the distances from the rotational axis to different points of the body. If a body consists of the finite number of particles then equation works. If, however, a rigid body has continuous structure with infinite number of particles, the equation has to be replaced with the integral definition I z r dm, (5.1.15) where integral is taken over the volume of the rigid body. 3. Calculating the moment of inertia. The moment of inertia can be calculated according to equation for a system of particles and according to equation for majority of the rigid bodies. The results for common bodies and given rotational axes are presented in Table 10- in the book. The question then becomes how the choice of rotational axis affects the moment of inertia. Suppose we already know this quantity I com for the axis of rotation that extends through the body's center of mass. This is the case for all the bodies in Table 10-. In fact, we have not discussed the definition of center of mass yet. We will do this in details later in the semester. Let me just briefly mention that position for the center of mass of the rigid body is defined as z rdm rcom, (5.1.16) m

6 where integral is taken over body's volume, m is the total mass of the body and r is the position vector of the mass element dm. For all symmetrical bodies considered in Table 10-, the position of the center of mass is matched with geometrical centers of the bodies. If we want to find the moment of inertia, I, for another axis, which is parallel to the original axis going through the center of mass and the distance between two axes is h then I I mh. (5.1.17) com The last equation is known as the parallel-axis theorem. To prove it, let us use the definition of the moment of inertia We shall place the origin of coordinate system to the body's center of mass defined by equation We will only need to consider the cross section of the body perpendicular to both axes and going through the body's center of mass. Let this plane be the xy plane of our coordinate system. We shall mark a and b to be the x, y coordinates of the new axis of rotation at this plane. This means that according to Pythagorean theorem h a b and b g b g for any element dm having coordinates x and y, so r x a y b I r dm x a y b dm x xa a y yb b dm I x y xa yb h dm x y dm a xdm b ydm h dm com mh, where we have made use of the fact that the origin was placed to the center of mass, so z z z d i, and. Icom x y dm xdm 0 ydm 0 4. Rolling One of the most common examples of the rigid body s rotation in the everyday life is rolling of the bicycle's wheel or the car's wheel along the road. In fact rolling is not a pure rotation, because the wheels are not only rotating but also performing translational motion along the surface of the road. Let us consider this type of motion in details. We shall consider a wheel rolling smoothly (without slipping) along the road. It participates in both translational and rotational motions. Let the center of mass O (see the picture) of this wheel be moving forward along the road with constant speed v com.

7 Point P where the wheel contacts the ground should also move with speed v com, so it can stay exactly below point O. This means that for time t this point P moves along the road for a distance s vcomt. At the same time, since this wheel moves without slipping, the point of the wheel which originally was at the ground rotates for the distance of the circular arc of the same length s R. Here is the angle for which this wheel has turned and R is the radius of the wheel. Taking time derivative of this equation one has ds dt b g b g d vcomt d R d vcom dt dt dt R R, so vcom R. (5.1.18) The linear velocity of wheel's rotation is the same as the linear velocity of wheel's translation. Every point of the wheel participates in both motions. The net velocity of each point can be calculated as the vector sum of both velocities. The center of the wheel O is not rotating; it only has velocity v com of translation. The point at the bottom of the wheel moves forward with v com due to translation and backward with - v com due to rotation, so its net velocity is zero. The point T at the top of the wheel moves forward with v com due to rotation as well as due to translation, so its total velocity is v com. Rolling can also be considered as pure rotation around the axis going through the point P which will give the same results. Considering rolling of the wheel, we assumed that it moves with constant speed. However, it is not always the case. In fact, the car or the bicycle can accelerate along the road with acceleration a com for its center of mass. In this case the wheel also accelerates and has the angular acceleration, which can be related to a com by taking the derivative of equation as

8 a com dvcom d dt dt R R, (5.1.19) where is the angular acceleration of this wheel. For a wheel to have angular acceleration the nonzero net torque is needed. This torque comes from the force of friction acting at point P. Since equations and are only valid in the case of smooth rolling without slipping, this means that friction force acting on the wheel at point P is, in fact, the force of the static friction, not of kinetic friction. This force opposes the wheel's tendency to slide. The direction of this frictional force depends on whether the wheel is accelerating or slowing down. It is the same direction as the direction of acceleration a com. Example Let us consider the situation similar to the example of the block sliding down along the inclined plan. The only difference is that we will replace the block of mass m by the rolling uniform body of the mass m. The question stays the same, to find acceleration of this body rolling down the ramp. The set of equations which we have obtained from the Newton's second law stays valid ma mg sin F, 0 N mg cos. f (5.1.0) However, the force of friction in these equations is no longer defined by the equation Ff kn, because it is not kinetic friction any longer. To find this force we need to use Newton's second law for rotation. Force F f is the only force in this problem, which has nonzero torque, because other forces are applied either at the pivot point as gravitational force or along the radius as the normal force, so we have

9 f RF F f I, f I, I Ia, R R (5.1.1) where I is the moment of inertia for this body, R is the radius of the body and is the angular acceleration. So the last equation gives Ia ma mg sin, R Fm I I R K J a mg sin, HG a mg sin m I R F I HG K J g sin. I 1 mr