PH 1-3A Fall 009 ROTATION Lectures 16-17 Chapter 10 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1
Chapter 10 Rotation In this chapter we will study the rotational motion of rigid bodies about a fixed axis. To describe this type of motion we will introduce the following new concepts: -Angular displacement -Average and instantaneous angular velocity (symbol: ω ) -Average and instantaneous angular acceleration (symbol: α ) -Rotational inertia also known as moment of inertia (symbol I ) -Torque (symbol τ ) We will also calculate the kinetic energy associated with rotation, write Newton s second law for rotational motion, and introduce the work-kinetic energy for rotational motion
The Rotational Variables In this chapter we will study the rotational motion of rigid bodies about fixed axes. A rigid body is defined as one that can rotate with all its parts locked together and without any change of its shape. A fixed axis means that the object rotates about an axis that does not move. We can describe the motion of a rigid body rotating about a fixed axis by specifying just one parameter. Consider the rigid body of the figure. We take the the z-axis to be the fixed axis of rotation. We define a reference line which h is fixed in the rigid id body and is perpendicular to the rotational ti axis. A top view is shown in the lower picture. The angular position of the reference line at any time t is defined by the angle θ(t) that the reference lines makes with the position i at t = 0. The angle θ(t) also defines the position i of all the points on the rigid body because all the points are locked as they rotate. The angle θ is related to the arc length s traveled by a point at a distance r from the axis via s the equation: θ = Note: The angle θ is measured in radians3 r
t Angular Displacement In the picture we show the reference line at a time t1 and at a later time t. Between t1 and t the body undergoes t 1 an angular displacement Δ θ = θ θ1. All the points of the rigid body have the same angular displacement because they rotate locked together. Angular Velocity We define as average angular velocity for the time interval t, t the ratio: ω avg θ θ Δθ t t Δ t 1 = = 1 ( ) The SI unit for angular velocity is radians/second Δθ We define as the instantaneous angular velocity the limit of as Δt 0 Δt Δθ dθθ ω = lim This is the definition of the first derivative with t ω = Δ t 0 Δt dt Algerbraic sign of angular frequency: If a rigid body rotates counterclockwise (CCW) ω has a positive sign. If on the other hand the rotation is clockwise (CW) 4 ω has a negative sign 1
t ω ω 1 t 1 Angular Acceleration If the angular velocity of a rotating rigid object changes with time we can describe the time rate of change of ω by defining the angular aceleration In the figure we show the reference line at a time t and at a later time t. 1 The angular velocity of the rotating body is equal to ω at t and ω at t. 1 1 We define as average angular acceleration fo ( ) We define as average angular acceleration fo r the time interval t, t the ratio: α avg ω ω1 Δω = = t t Δt 1 1 The SI unit for angular velocity is radians/second Δω We define as the instantaneous angular acceleration the limit of as Δt 0 Δt Δω α = lim This is the definition iti of the first tderivative with ithtt dωω Δ t 0 Δ t α = dt 5
Angular Velocity Vector For rotations of rigid bodies about a fixed axis we can describe accurately the angular velocity by asigning an algebraic sigh. Positive for counterclockwise rotation and negative for clockwise rotation We can actually use the vector notation to describe rotational motion which is more complicated. The angular velocity vector is defined as follows: The directi o n of r ω is along the rotation axis. The sense of r ω is defined by the right hand rule (RHL) Right hand rule: Curl the right hand so that the fingers point in the direction r of the rotation. The thumb of the right hand gives the sense of ω 6
Rotation with Constant Angular Acceleration When the angular acceleration α is constant we can derive simple expressions that give us the angular velocity ω and the angular position θ as function of time We could derive these equations in the same way we did in chapter. Instead we will simply write the solutions by exploiting the analogy between translational and rotational motion using the following correspondance between the two motions Translational Motion Rotational Motion x θ v ω a α v= v + at 0 ω = ω + αt 0 (eqs.1) at αt x = x o + v ot+ θ = θ + ω ot + (eqs.) (q v v = a( x x ) ω ω = α( θ θ ) ( eqs.3) 7 o o o o
O θ A s Relating the Linear and Angular Variables Consider a point P on a rigid body rotating about a fixed axis. At t = 0 the reference line which connects the origin O with point P is on the x-axis (point A) During the time interval t point P moves along arc AP and covers a distance s. At the same time the reference line OP rotates by an angle θ. Relation between angular velocity and speed The arc length s and the angle θ are connected by the equation: s ( θ ) ds d r dθ = rθ where r is the distance OP. The speed of point P v= = = r dt dt dt v = rω The period T of revolution is given by: T = π 1 T = ω = π f ω f T circumference π r πr π = = = = speed v ωr ω 8
O r The Acceleration The acceleration of point P is a vector that has two components. A "radial" componet along the radius and pointing towards point O. We have enountered this component in chapter 4where we called it "centripetal" acceleration. Its magnitude is: a r v = = ω r r The second component is along the tangent to the circular path of P and is thus known as the "tangential" component. Its magnitude is: ( ωr ) dv d dω at = = = r = rα dt dt dt at = rα The magnitude of the acceleration vector is : a = at + ar 9
Example. Motion with constant angular acceleration. 10
Rotation with constant angular acceleration. 11
The equations of rotational kinematics. 1
The equations of kinematics. 13
Angular variables and tangential variables. 14
Angular variables and tangential variables. 15
Centripetal acceleration and tangential acceleration. 16
v r i Kinetic Energy of Rotation Consider the rotating rigid body shown in the figure. mi We divide id the body into parts of masses m r 1, m, m3,..., mi,... i O The part (or "element") at P has an index i and mass mi The kinetic energy of rotation is the sum if the kinetic 1 1 1 energies of the parts K = mv 1 1+ mv+ m3v3+... 1 1 K = mv The speed of the -th element ( ) i i i vi= ωri K = mi ωri i i 1 1 K = mr ω = Iω The term I = mr is known as i i i i i i rotational inertia or moment of inertia about the axis of rotation. The axis of rotation must be specified because the value of I for a rigid body depends on its mass, its shape as well as on the position of the rotation axis. The rotational inertia of an object describes howthe mass is distributed about the rotation axis I = mr 1 i i I = r dm K = Iω 17 i
An automobile of mass 1400kg has wheels 0.75m in diameter weighing 7kg each. Taking into account the rotational kinetic energy of the wheels about their axes, what is the total kinetic energy of the automobile traveling at 80km/h? What percent of kinetic energy belongs to the rotational motion of the wheels about their axes? Pretend that each wheel has a mass distribution equivalent to that of a uniform disk. 18
In the table below we list the rotational inertias for some rigid bodies I = r dm 19
Calculating the Rotational Inertia = The rotational inertia I mr i i This expression is useful for a rigid body that i has a discreet disstribution of mass. For a continuous distribution of mass the sum becomes an integral I = r dm Parallel-Axis Theorem We saw earlier that I depends on the position of the rotation axis For a new axis we must recalculate the integral for I. A simpler method takes advantage of the parallel-axis theorem Consider the rigid body of mass M shown in the figure. We assume that we know the rotational inertia about a rotation axis that passes through the center of mass O and is perpendicular t o the page. The rotational inertia I about an axis parallel to the axis through O that passes through h point P, a distance h from O is given by the equation: I I Mh 0 = com + I com
A Proof of the Parallel-Axis Theorem We take the origin O to coincide with the center of mass of the rigid body shown in the figure. We assume that we know the rotational inertia for an axis that is perpendicular to the page and passes Icom through O. We wish to calculate the rotational ineria I about a new axis perpendicular ( ab) ( ) to the page and passes through point P with coodrinates,. Consider an element of mass dm at point A with coordinates x, y. The distance r between points A and P is: ( ) ( ) r = x a + y b Rotational Inertia about P: I = r dm x a y b = + dm ( ) ( ) I = ( x + y ) dm a xdm b+ ( a + b ) dm The second and third integrals are zero. The first integ ral is com. The term ( ) Thus the fourth integral is equal to h dm = Mh I Icom Mh I a + b = h = + 1
According to spectroscopic measurements, the moment of inertia of an oxygen molecule about an axis through the center of mass and perpendicular to the line joining the atoms is 1.95x10-46 kg m. The mass of an oxygen atom is.66x10-6 g. What is the distance between atoms?
Torque In fig.a we show a body which can rotate about an axis through point O under the action of a force F r applied at point P a distance r from O. In fig.b we resolve F r into two componets, rdial a and tangential. The radial component F r cannot cause any rotation because it acts along a line that passes through O. The tangential component Ft = Fsin φ on the other hand causes the rotation of the r object about O. The ability of F to rotate the body depends on the magnitude Ft and also on the distance r between points P and O. Thus we define as torque τ = rft = rf si n φ = r F The distance r is known as the moment arm and it is the perpendicular distance between point O and the vector F r The algebraic sign of the torque is asigned as follows: If a force F r tends to rotate an object in the counterclockwise direction the sign is positive. If a force F r tends to rotate t an object in the clockwise direction the sign is negative. τ = rf 3
Newton's Second Law for Rotation For translational motion Newton's second law connects the force acting on a particle with the resulting acceleration There is a similar relationship between the torque of a force applied on a rigid object and the resulting angular acceleration This equation is known as Newton's second law for rotation. We will explore this law by studying a simple body which consists of a point mass m at the end r of a massless rod of length r. A force F is applied on the particle and rotates r the system about an axis at the origin. As we did earlier, we resolve F into a tangential and a radial component. The tangential component is responsible for the rotation. tti We first apply Newton's second dlaw for Ft. Ft = mat (eqs.1) The torque τ acting on the particle is: τ = Fr (eqs.) We eliminate F between equations 1 and : τ t ( ) ( ) τ = ma r = m α r r = mr α = I α t = Iα (compare with: F = ma ) 4 t
O i r i 1 3 Newton's Second Law for Rotation We have derived Newton's second law for rotation for a special case. A rigid body which consists of a point mass m at the end of a massless rod of length r. We will now derive the same equation for a general case. Consider the rod-like object shown in the figure which can rotate about an axis through point O undet the action of a net torque τ net. We divide the body into parts or "elements" and label them. The elements have masses m1, m, m3,..., mn and they are located at distances r1, r, r3,..., rn from O. We apply Newton's second law for rotation to each element: τ1 = I1α (eqs.1), τ = Iα (eqs.), τ3 = I3α (eqs.3), etc. If we add all these equations we get: τ + τ + τ +... + τ = I + I + I +... + I α. Here I = mr is the rotational linertiai ( ) 1 3 n 1 3 n i i i of the i-th element. The sum τ1+ τ + τ3+... + τn is the net torque τnet applied. The sum I 1 + I + I 3 +... + In is the rotational inertia I of the body. Thus we end up with the equation: τ net = Iα 5
A block of mass m hangs from a string wrapped around a frictionless pulley of mass M and radius R. If the block descends from rest under the influence of gravity, what is the angular acceleration of the pulley? 6
Work and Rotational Kinetic Energy In chapter 7 we saw that if a force does work W on an object, this results in a change of its kinetic energy Δ K = W. In a similar way, when a torque does work W on a rotating rigid body, it changes its rotational kinetic energy by the same amount W = ΔK Consider the simple rigid body shown in the figure which consists of a mass m r at the end of a massless rod of length r. The force F does work dw = Frd t θ = τdθ The radial component F does zero work because it is at right angles to the motion. The work r f W = Frdθ = τdθ. By virtue of the work-kinetic energy theorem we have a change in kinetic energy W Δ K t θ θ i 1 1 1 1 Δ K = W = mv f mvi = mr ωf mr ωi = θ f 1 1 Δ K = Iω f Iωi W = τ d θ θ i 7
Conservation of Energy in Rotational Motion A meter stick is initially standing vertically on the floor. If it falls over, with what angular velocity will it hit the floor? Assume that the end in contact with the floor doesn t slip. 8
Power Power has been defined as the rate at which work is done by a force and in the case of rotational motion by a torque We saw that a torque τ produces work dw = τdθ as it rotates an object by an angle dθθ. dw d dθ P = = ( τdθ) = τ = τω (Compare with P= Fv) dt dt dt Below we summarize the results of the work-rotational kinetic energy theorem W θ f τd θ W = τ ( θ ) f θi For constant torque = ( ) For constant torque θ i 1 1 W =Δ K = Iω f Iω Work-Rotational Kinetic Energy Theorem i P = τω 9
Analogies between translational and rotational Motion Translational Motion 0 x v Rotational Motion θ ω a α v= v + at ω = ω + αt at αt x= xo + vot+ θ = θo + ωot + v v a x x ω ω α θ θ = ( ) ω ω = α ( θ θ ) v vo a x xo mv Iω K = K = m I F = ma τ = Iα F τ P = Fv P 0 o = τω o 30