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Chapter 8 Rotational Kinematics

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Chapter 8 Rotational Kinematics Goals for Chapter 8 Chapter 8 • To study angular velocity and angular acceleration. Rotational Kinematics • To examine rotation with constant angular acceleration. • To understand the relationship between linear and ang ular q uantities . • To determine the kinetic energy of rotation and the moment of inertia. • To study rotation about a moving axis. afs p53f09 L18 afs p53f09 L18 Rigid Bodies can Rotate around a Fixed Axis Rotational Motion and Angular Displacement In the simplest kind of rotation, points on a rigid object move on circular paths around an axis of rotation. The angle through which the object rotates is called the angular displacement. Δθ = θ − θo afs p53f09 L18 afs p53f09 L18 Definition of Angular Displacement Angular Displacement is measured in Radians When a rigid body rotates about a fixed axis, the angular Arc length s displacement is the angle swept out by a line passing θ (in radians) = = Radius r through any point on the body and intersecting the axis of rotation perpendicularly. By convention, the angular displacement is positive if it is counterclockwise and 2πr negative if it is clockwise. For a full revolution: θ = = 2π rad r 2π rad = 360o 2 π radians = 360o 360° 180° 1 rad = = = 57.3°. SI Unit of Angular Displacement: radian (rad) 2π π afs p53f09 L18 afs p53f09 L18 Example: Adjacent Synchronous Satellites Example: A Total Eclipse of the Sun Synchronous satellites are put into The diameter of the sun is about 400 times greater an orbit whose radius is 4.23×107m. than that of the moon. By coincidence, the sun is also about 400 times farther from the earth than is If the angular separation of the two the moon. satellites is 2.00 degrees, find the For an observer on the earth, compare the angle arc length that separates them. subtended by the moon to the angle subtended by Arc length s the sun and explain why this result leads to a total θ (in radians) = = solar eclipse. Radius r ⎛ 2π rad ⎞ 2.00deg ⎜ ⎟ = 0.0349 rad ⎝ 360deg ⎠ s = r θ = (4.23×107 m)()0.0349 rad =1.48×106 m (920 miles) afs p53f09 L18 afs p53f09 L18 Angular Velocity and Angular Acceleration Δθ = θ − θo Arc length s How do we describe the θ (in radians) = = Radius r rate at which the angular displacement is changing? s 400 smoon smoon sun = = = θmoon θsun = 400 r DEFINITIOOAON OF AVER AGE moon rmoon rsun ANGULAR VELOCITY Angular displacement The angles subtended by the sun and the moon are approximately Average angular velocity = equal. Therefore it is possible to have a total solar eclipse Elapsed time θ − θ Δθ ω = o = SI Unit of Angular Velocity: t − t o Δt radian per second (rad/s) afs p53f09 L18 afs p53f09 L18 Example: Gymnast on a High Bar (Instantaneous) Angular Velocity A gymnast on a high bar swings through two • The average angular velocity revolutions in a time of 1.90 s. (speed), ω, of a rotating rigid Find the average angular object is the ratio of the angular velocityof the gymnast. displacement to the time interval ⎛ 2π rad ⎞ θf − θi Δθ Δθ = −2.00 rev⎜ ⎟ ω = = ⎝ 1rev ⎠ t f − ti Δt = −12.6 rad (instantaneous) angular velocity Δθ −12.6 rad ω = ω = lim ω = lim 1.90 s Δt→0 Δt→0 Δt = −6.63rad s Radians per second is unit for ω, the angular velocity afs p53f09 L18 afs p53f09 L18 Angular Acceleration Example: A Jet Revving its Engines If we follow what we studied in linear As seen from the front of the motion, when velocity changes, we engine, the fan blades are call the rate of change --- acceleration rotating with an angular The average angular acceleration, speed of -110 rad/s. As the α, of a rotating rigid object is the plane takes off, the angular ratio of change in angular velocity velocity of the blades reaches to the time interval -330 rad/s in a time of 14 s. ω− ω Δω Find the angular acceleration, α = o = assuming it to be constant. t − t o Δt (− 330rad s)− ()−110rad s (instantaneous) SI Unit of Angular α = angular acceleration acceleration: radian per 14 s Δω second squared (rad/s2) 2 α = lim = −16rad s Δt→0 Δt afs p53f09 L18 afs p53f09 L18 Rotational Variables Arc Length Angular Position AlDilAngular Displacemen t AlAngular spee ddlitd and velocity afs p53f09 L18 afs p53f09 L18 Rotational Kinematics and Variables Sign convention for angular displacement and velocity Angular Displacement θ Arc length s θ (in radians) = = Radius r 2 π radians = 360o angular velocity Δθ Unit: rad / sec ω = lim ω = lim Δt→0 Δt→0 Δt angular acceleration Δω α = lim ω = lim Unit: rad / sec2 Δt→0 Δt→0 Δt afs p53f09 L18 afs p53f09 L18 Example: Rotation of a Compact Disc Recall the equations of ‘translational’ kinematics A compact disc (CD) rotates at high speed while a laser reads data for constant acceleration. encoded in a spiral pattern. The disk has radius R = 6.0 cm. When data are being read, it spins at 7200 rev/min. Five kinematic variables: a = const. What is the CD’s angular velocity ω in radians per 1. displacement, x second ? How much time is required for it to rotate o 2. acceleration (constant), a through 90 ? If it starts from rest and reaches full v = vo + at speed in 4.0 s, what is its average angular acceleration? 3. final velocity (at time t), v 2 Δθ 7200rev2 πrad rad 4. initial velocity, vo 1 ω= ω = =754 x = vo t + 2 at Δt 60s rev s 5. elapsed time, t Δ θ π 2 rad Δt = = = 2.1×10−3 s ω 754 rad s 1 2 2 Δω ω−0 754 rad s rad a = = = =189 x = 2 (vo + v) t v = vo + 2ax av Δt Δ t 4.0s s2 afs p53f09 L18 afs p53f09 L18 The Equations of Rotational Kinematics The Equations of Rotational Kinematics angular acceleration: α = const. ANGULAR ACCELERATION ANGULAR VELOCITY ω = ωo + αt TIME 1 2 θ = ωot + 2 αt ANGULAR DISPLACEMENT 1 2 2 θ = 2 ()ωo + ω t ω = ωo + 2α θ afs p53f09 L18 afs p53f09 L18 Example: Rotation of a Bicycle Wheel The angular velocity of the rear wheel of a stationary exercise bike is 4.00 rad/s at time t=0, and its angular acceleration is rad 2 constant and equal to 2.00 rad/s . A particular spoke coincides θo = 0 ωo = 4.00 with the x-axis at t=0. What angle does this spoke make with the s +x-axis at time t=3.00 s ? What is the wheel’s angular velocity at rad α = 2.00 t = 3.00 s this time ? s2 1 θ = θ + ω t + α t 2 o o 2 t = 3.00 s ⇒ θ = 21.0 rad ω = ωo + αt rad rad rad t = 3.00 s ⇒ ω = 4.0 + 2 ⋅ 3 = 10 .0 s s s afs p53f09 L18 afs p53f09 L18 The Equations of Rotational Kinematics Example: Blending with a Blender The blades are whirling with an angular velocity of Reasoning Strategy +375 rad/s when the “puree” button is pushed in. 1. Make a drawing. When the “blend” button is pushed, the blades accelerate and reach a greater angular velocity after 2. Decide which directions are to be called positive (+) and the blades have rotated through an angular negative (-). displacement of +44.0 rad. The angular acceleration 3. Write down the values that are given for any of the five has a constant value of +1740 rad/s2. kinematic variables. Find the final angular velocity of the blades. 4. Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation. 2 2 ω = ω2 + 2αθ ω = ωo + 2α θ o 5. When the motion is divided into segments, remember that the final angular velocity of one segment is the initial velocity for the next. ω = ()375rad s 2 + 2()1740 rad s2 ()44.0rad 6. Keep in mind that there may be two possible answers to a kinematics problem. = +542 rad s afs p53f09 L18 afs p53f09 L18 Angular Variables and Tangential Variables Centripetal Acceleration and Angular Acceleration v − v a = T To T t ()(rω − rω ) ω− ωo = o = r t t ω− ω α = o t θ ω = a = r α (α in rad/s2 ) t T s r θ ⎛ θ ⎞ vT = = = r ⎜ ⎟ 2 2 t t ⎝ t ⎠ v T (rω) 2 (ω in rad/s) a c = = = rω vT = r ω (ω in rad/s) r r afs p53f09 L18 afs p53f09 L18 Relationship between Linear and Angular Quantities Example: A Helicopter Blade A helicopter blade has an angular speed of 6.50 rev/s and an Δs 2 • Displacement Δθ = angular acceleration of 1.30 rev/s . For point 1 on the blade, find r the magnitude of (a) the tangential speed and (b) the tangential acceleration. ⎛ rev ⎞ ⎛ 2π rad ⎞ ω = 6.50 = 40.8 rad s v ⎜ ⎟ ⎜ ⎟ • Speed ω = ⎝ s ⎠ ⎝ 1 rev ⎠ ⎛ rev ⎞ ⎛ 2π rad ⎞ r α = ⎜1.30 ⎟ ⎜ ⎟ = 8.17 rad s2 ⎝ s2 ⎠ ⎝ 1 rev ⎠ 2 a T • Acceleration α = 2 v a = a = ω r = vT = r ω r C rad r VT = (3.00 m) ()40.8rad s =122m s When a rigid object rotates about a fixed axis, every portion of the object has the same angular speed and the same angular acceleration a T = r α i.e. θ, ω, and α are not dependent upon r, distance form hub or axis of rotation 2 2 aT = (3.00 m) (8.17 rad s )= 24.5m s afs p53f09 L18 afs p53f09 L18 Example: Acceleration of Your Car on a Circular Roadway c) To avoid a rear-end collision with Suppose you are driving a car in a counterclockwise a vehicle ahead, you apply the direction on a circular road whose radius is r = 390 brakes and reduce your angular m.
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