Physics 141 Rotational Motion 2 Page 1. Rotational Motion 2

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1 Physics 141 Rotational Motion 2 Page 1 Rotational Motion 2 Right handers, go over there, left handers over here. The rest of you, come with me.! Yogi Berra Torque Motion of a rigid body, like motion of any system of particles, is changed by the effects of external forces. A rigid body's motion consists of motion of the CM plus rotation about the CM; external forces can change one or both of these parts of the motion. Shown is a rigid body, with an external force F applied to it at a point specified by the position vector r, relative to the CM as origin. At first we take the force vector F to lie in the plane of the drawing. This vector can be broken up into a component along r ( F r ) and a component perpendicular to r ( F ). Acting alone, the radial component F r would make the CM accelerate to the right along the line of r, but it would not produce any rotation about the CM. Now suppose there is a fixed axle passing through the body, perpendicular to the page and passing through the CM. Then normal forces exerted on the body by the axle would cancel the effect of F r and the body would not move at all. We will ignore this component. On the other hand, F would make the body rotate about the axle. It is this rotational effect that we are interested in here. As the body turns counter-clockwise through angle dθ during time dt, the point at which F is applied moves through distance ds = r dθ, so the work done by it is rf dθ. This infinitesimal work causes an infinitesimal increase in the rotational kinetic energy: ( ). rf dθ = d 1 2 Iω 2 The quantity rf, acting as the body rotates through angle dθ, produces an infinitesimal change in the rotational kinetic energy. This suggests that the quantity rf is a useful measure of the effectiveness of the force in producing angular motion, i.e., rotation. This quantity is called the torque of the force, denoted by the Greek letter τ: τ = rf. CM Torque is a vector quantity. We have introduced the relevant component of the torque, in the case of a rigid body constrained to rotate about a fixed axle, and with applied force perpendicular to that axle. (In such a case the torque has only one component.) A more general definition will be given later. r F F r F

2 Physics 141 Rotational Motion 2 Page 2 Since torque involves the distance r, it depends on the location of the reference point about which the body rotates. For the example considered here, it is the point where the axle passes through the plane of the diagram. (We have taken that point to be the CM for simplicity, but that is only one choice.) To specify a torque requires specifying the reference point about which it acts. Moment arm and torque Shown is the same situation discussed above. We see that since F = F sinθ the torque s magnitude is given by τ = rf sinθ. But this can also be written as τ = r F, where r = r sinθ. The distance r is called the moment arm of the force. CM r r F θ F F r Moment arm of a force The moment arm of a force in producing a torque is the perpendicular distance from the reference point to the line along which the force acts. In terms of this definition: The magnitude of the torque is the product of the moment arm and the magnitude of the force. If a force acts directly toward or away from the reference point, its moment arm is zero and the force produces no torque about that reference point. The moment arm is a maximum when F r, in which case the moment arm is simply r. In the case discussed so far, torques produce either clockwise or counter-clockwise rotation about the axle. One can choose which of these directions to count as positive, and a torque that will produce positive rotation is counted as a positive torque. The three-dimensional aspects of the direction of the torque vector will be discussed later. Relation between torque and angular acceleration It is relatively easy to compute the torques, about a given point, due to external forces on a rigid body. It is much more subtle to relate these torques to the actual motion of the body, if it is not constrained by a fixed axle. Many of these subtleties are beyond the scope of this course. Even if the body is constrained to rotate about a fixed axle, there can be some subtle aspects, as will be discussed below. Only if the body is symmetric about its rotation axis is the situation easy to analyze. Here symmetric means that each point mass in the body has an identical mirror image point mass with the same mass at the same distance directly on the other side of the axis. For such a body the symmetry axis automatically passes through the CM. Bodies with uniform mass density which are figures of revolution about the axis (wheels, pulleys, etc.) have such symmetry. For symmetric bodies, as will be shown later, there is a simple and useful relation between the net torque due to external forces and the angular acceleration of the body about the fixed axis:

3 Physics 141 Rotational Motion 2 Page 3 Torque and angular acceleration (symmetric body, fixed axis) τ tot = Iα This relation is one case of what is sometimes called the rotational 2 nd law. Rolling An important example of a symmetric body rotating about its symmetry axis is the rolling motion of a uniform figure of revolution, such as a wheel or a ball, on a surface. Rolling is motion in which the point of contact between the moving object and the surface over which it is moving is always instantaneously at rest. That is, the moving body does not slide on the surface. To see the consequences of this requirement, it is useful to consider the motion in the frame moving with the CM of the body. In that frame, the motion of the rolling body is nothing but rotation about its symmetry axis, while the surface passes backwards underneath it with speed v CM. A point on the outer rim of the body will move around the rim with speed v = Rω, where R is the radius. If the point of contact is always to be at rest relative to the surface, the backward speed of the lowest point on the rim must equal the speed of the backward moving surface. This gives the rolling condition: Rolling condition v CM = Rω If the body accelerates while still not slipping, then this condition must hold at all times, and we must have a CM = Rα. An example is a wheel rolling down an incline. For a rolling symmetric body the kinetic energy is Kinetic energy of rolling K = 1 2 Mv CM Iω 2 This formula, combined with the rolling condition, has many applications. Static equilibrium of a rigid body A rigid body in static equilibrium is completely at rest and remains so. The CM remains at rest, and there is no rotation about the CM. These things can happen only if both of the following are true: The total external force is zero. The total torque about the CM is zero. The general conditions for static equilibrium are thus: Static Equilibrium Translational: F ext tot = 0. Rotational: τ ext tot = 0.

4 Physics 141 Rotational Motion 2 Page 4 These are both conditions on vectors, so they must hold for all components. One can easily show that any point can be used for the reference point for the torques (not just the CM) as long as the total force is zero. In the cases treated in this course, the forces usually all lie in a plane, so the condition on the forces involves two component equations. Any torque about a point in that plane will have only a component perpendicular to the plane, so the condition on the torques gives only one equation. Only situations with three or fewer unknowns can be completely determined by these conditions. Stress and strain We have usually treated forces as though they act at a single point. This is at best an approximation. Generally forces are distributed over parts of the object. Gravity, for example, acts individually on every particle in the body. The total gravitational force is the sum of these individual forces. For a small object near the earth's surface, the force on each particle is simply mg. If g is the same at the location of all the particles, the total gravitational torque about the CM of the body is easily shown to be zero. For that reason we say that gravity acts effectively at the CM of the body. This is not true of large bodies like the earth itself, acted on by gravity from other bodies like the sun and moon, since g is not the same at the location of all the particles. One consequence of this is tidal forces and torques, which will be discussed later. The normal force exerted by a surface on a body acts at all the atoms and molecules of the common interface. Friction similarly acts at all points on the interface. To discuss the distribution of a force over a surface, one introduces the concept of force per unit area, which is called the stress. Stresses are divided into three categories: Tension acts to pull particles at the surface directly away from the body. Compression acts to push particles at the surface directly into the body. Shear acts to move particles at the surface parallel to that surface. Tensile and compressive stresses involve forces normal to the surface of the body, while shear stresses involve forces along the surface. Stresses generally produce (at least momentary) deformations of the body. These are called strains. If the deformation is small and not permanent, then we have a situation like Hooke's law for springs. The strain is (approximately) linearly proportional to the stress, and the constant of proportionality is called an elastic modulus. In the case of tension or compression the fractional change in the dimension of the object perpendicular to the interface is related to the tensile or compressive stress by Young's modulus:! Y = Stress Strain = F /A ΔL/L. Here F/A is the force per unit area normal to the interface, and the strain is the fractional change in the dimension normal to the interface.

5 Physics 141 Rotational Motion 2 Page 5 In the case of shear, the strain is defined to be the ratio of the lateral deformation (Δx) to the dimension perpendicular to the interface (h). The corresponding modulus is called the shear modulus:! S = F /A Δx/h. In this case, F/A is the ratio of the force along the interface to the area of the interface. An object may be subject to forces acting at all points on its surface. If these forces are everywhere normal to the surface, the effect is to compress (or expand) the volume occupied by the object. In this case the force per unit area is called pressure:! P = F /A. P is positive if the force on the body is inward, i.e., compressive. If the pressure increases by ΔP, the volume V occupied by the body will decrease by ΔV. The relationship is given by the bulk modulus:! B = ΔP ΔV /V. We will return to pressure when we discuss fluids.

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