Presentation on theme: "Rotational Motion Angular Quantities Angular Kinematics Rolling Motion."— Presentation transcript:
Rotational Motion Angular Quantities Angular Kinematics Rolling Motion
Chapter 8 Rotational Motion We consider only rigid (non-squishy) bodies in this chapter. We have already studied translational motion in considerable detail. We begin this chapter by considering purely rotational motion (the center of mass does not change its xyz coordinates)… We have already studied translational motion in considerable detail. We now know that the translational motion we have been describing is the motion of the center of mass. We begin this chapter by considering purely rotational motion (the center of mass does not change its xyz coordinates)… but soon we will consider objects which both rotate and translate.
8.1 Angular Quantities Look at a point P somewhere on a rotating circular disc. For rotational motion, we specify angles in radians instead of degrees. P is a distance r from the center of the disc. Choose the x-axis to be horizontal. Then the line from the center of the disc to P makes an angle with the x-axis. =ℓ/r, where ℓ is the length of the arc from the x-axis to P. P x O ℓ r
P x O ℓ If the angle is 360 degrees, the arc length is 2r, so 2 radians are equal to 360 degrees. This is how I remember the conversion factor between degrees and radians. Strictly speaking, radians are not a unit (angles are unitless). When you studied kinematics, you defined an object’s motion by specifying its position, velocity, and acceleration. Now we are about to study angular kinematics. We will define an object’s angular motion by specifying its angle (rotational position relative to some axis), angular velocity, and angular acceleration.
If we call the z axis the axis of rotation, angular velocity is defined by avg =/t, where is the angular displacement of a rotating object in the time t. Instantaneous angular velocity is =d/dt. “Wait a minute! Velocity is a vector! What is the direction of ?” Consider our rotating wheel. Here’s a point and its instantaneous linear velocity. Another.Another. And another. How do we define a unique direction for the angular velocity?
If we put the x and y axes in the plane of the wheel, then the wheel is rotating about an axis perpendicular to this plane. We usually call this the z-axis. z To remind you that angular velocity has a direction, I’ll write z,avg =/t and z =d/dt. The direction of is along the z- axis, perpendicular to the wheel, and is given by the right hand rule. Our rules for vectors apply. You get to choose the direction of the z-axis. Whether z is positive or negative depends on the direction of rotation. this symbol indicates an arrow coming out of the screen
If this business about direction is confusing, don’t worry. It is not particularly important this semester. You’ll learn to love the right hand rule if you are lucky enough to take the second semester of algebra-based physics! Note that all points on a rigid body rotate with the same angular velocity (they go around once in the same time). What about the tangential velocities of different points? We’ll get to that question in a minute.
Angular velocity z,avg =/t and z =d/dt. You can probably figure out how angular acceleration is defined. P x ℓ z z,avg = z /t, where z is the change in angular velocity of a rotating object in the time t. Instantaneous angular acceleration is z =d z /dt. The subscript z on and emphasizes the rotation is relative to some axis, which we typically label “z.” I should really put a subscript z on because it is also measured relative to the z axis. To be consistent with Physics 23, I will leave it off.
Because points on a rotating object also have an instantaneous linear motion, linear and angular motion must be connected. As your book shows, v tangential =r z and a tangential =r z. Note that z and z are the same for all points on a rotating rigid body, but v tang and a tang are not. zz Notation: v tangential =v tang =v .. r v tang , , v and a are all magnitudes of vectors! z and z are vector components!
We often use frequency and period of rotation: f=/2 and T=1/f. Acceleration has both tangential and radial components. We know from before that a r =v 2 /r. Total acceleration is a=a tangential +a r. Example 8.3. What are the linear speed and acceleration of a child seated 1.2 m from the center of a steadily rotating merry-go-round that makes one complete revolution in 4.0 s? z = 2/4 s -1 v tang = r z = (1.2 m) (/2 s -1 ) v tang r z v tang = 0.6 m/s z = 0 (the angular velocity is not changing)
a tang = r z = 0 v tang r z a radial = a r = (0.6 m/s) 2 / (1.2 m) a r = 2.96 m/s 2 The total acceleration is the vector sum of the radial and tangential accelerations: a = 2.96 m/s 2, towards center of merry-go-round New OSE’s introduced in this section: z,avg =/t z,avg = z /t v tang =r z a tang =r z You can also use a radial = a R =R z 2
8.2 Kinematic Equations for Uniformly Accelerated Rotational Motion Remember our equations of kinematics from chapter 2? Analogous equations hold for rotational motion. linear angular v x =v 0x +a x t z = 0z + z t x=x 0 +v 0x t+½a x t 2 θ=θ 0 + 0z t+½ z t 2 v x 2 =v 0x 2 +2a x (x-x 0 ) z 2 = z 2 +2 z (θ-θ 0 ) new OSE’s
Example 8.6. Through how many revolutions did the centrifuge motor of example 8.5 turn during its acceleration period? Assume constant angular acceleration. In example 8.5, a centrifuge motor goes from rest to 20,000 rpm in 5 minutes. First we need to calculate , as in example 8.5. z,avg = z /t z,avg =( f - i ) / (t f -t i ) f =(20000rev/min)(min/60s)(2 radians/rev) t f =(5 min)(60 s/min) z,avg =7.0 rad/s 2 00
Next calculate the angle through which the centrifuge motor turns. θ=θ 0 + 0z t+½ z t 2 00 θ=½ (7.0 rad/s 2 ) (300 s) 2 θ=3.15x10 5 radians There are 2 radians in each revolution so the number of revolutions, N, is N=(3.15x10 5 radians)(revolution/2 radians) N=5x10 4 revolutions). I chose to do the calculations numerically rather than symbolically for practice with angular conversions.
8.3 Rolling Motion Many rotational motion situations involve rolling objects. Rolling without slipping involves both rotation and translation. Friction between the rolling object and the surface it rolls on is static, because the rolling object’s contact point with the surface is always instantaneously at rest. the illustration of in this diagram is misleading; the direction of would actually be into the screen this point on the wheel is instantaneously at rest if the wheel does not slip (slide)
Your text discusses the rolling motion of a bicycle wheel by changing reference frames; the center of mass of the wheel is taken to be at rest with the ground moving past it. We skipped the sections on relative motion, so I won’t expect you to perform a detailed analysis of rolling motion by switching reference frames. However, you should be able to calculate how many revolutions the bicycle wheel makes in example 8.7.
The point on the rolling object in contact with the “ground” is instantaneously at rest. v bot =0 The center of the wheel moves with the speed of the center of mass. v cm The point at the “top” of the rolling wheel moves with a speed twice the center of mass. v top =2v cm v cm The two side points at the level of the center of mass move vertically with the speed of the center of mass. Here are some things you need to know about rolling without slipping.
What is important for us is that if an object rolls without slipping, we can use our OSE’s for z and z, using the translational speed of the object. v bot =0 v cm v top =2v cm v cm v CM =r z a CM =r z These relationships will be valuable when we study the kinetic energy of rotating objects.
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