Chapter 3.5. Uniform Circular Motion

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Transcription:

Chapter 3.5 Uniform Circular Motion

3.5 Uniform Circular Motion DEFINITION OF UNIFORM CIRCULAR MOTION Uniform circular motion is the motion of an object traveling at a constant speed on a circular path. Circumference of the circle is 2! r.

3.5 Uniform Circular Motion The time it takes the object to travel once around the circle is T (a.k.a. the period) Speed around the circle is, v = 2! r T.

3.5 Uniform Circular Motion Example: A Tire-Balancing Machine The wheel of a car has a radius of 0.29m and it being rotated at 830 revolutions per minute on a tire-balancing machine. Determine the speed at which the outer edge of the wheel is moving.

3.5 Centripetal Acceleration In uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant. Point P Point O Angle between point O and point P is the same as between! v 0 and! v. t = 0 Since velocity vector changes direction Acceleration vector is NOT ZERO. v v a = 0 t Need to understand:! v!! v 0 NOTE:! v!! v 0 and! a point in toward center of circle! v v 0!! v 0! v! v0! v! v v 0

3.5 Centripetal Acceleration In uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant. Point P Point O a c Angle between point O and point P is the same as between! v 0 and! v. t = 0 Since velocity vector changes direction Acceleration vector is NOT ZERO. v v a = 0 t Need to understand:! v!! v 0 NOTE:! v!! v 0 and! a point in toward center of circle! v v 0!! v 0! v! v!! v0 v v0

3.5 Centripetal Acceleration Compare geometry of velocity vectors and the portion of the circle.! =!v v t = 0! = vt r Magnitudes!v v = vt r a c = Δv t = v2 r vt a c = v2 r

3.5 Centripetal Acceleration The direction of the centripetal acceleration is towards the center of the circle; in the same direction as the change in velocity. Point O a c t = 0! v!! v 0 v v0 Point P a c = v2 r Centripetal acceleration vector points inward at ALL points on the circle

3.5 Centripetal Acceleration Conceptual Example: Which Way Will the Object Go? An object ( ) is in uniform circular motion. At point O it is released from its circular path. Does the object move along the (A) Straight path between O and A or (B) Along the circular arc between points O and P?

3.5 Centripetal Acceleration Example: The Effect of Radius on Centripetal Acceleration The bobsled track contains turns with radii of 33 m and 24 m. Match the acceleration vector directions below to the points A,B,C,D. A 1 2 3 4 B A B C D D C

3.5 Centripetal Acceleration Example: The Effect of Radius on Centripetal Acceleration The bobsled track contains turns with radii of 33 m and 24 m. Match the acceleration vector directions below to the points A,B,C,D. A 4 1 2 3 4 B 3 A 4 B 3 C 1 D 2 1 2 D C

3.5 Centripetal Acceleration a C = v 2 Find the centripetal acceleration at each turn for a speed of 34 m/s. Express answers as multiples of r ( a C = 34m s ) 2 33 m = 35m s2 = 3.6g ( a C = 34m s ) 2 24 m = 48m s2 = 4.9g

Clicker Question 3.5.1 A car is moving counter-clockwise around a circular section of road at constant speed. What are the directions of its velocity and acceleration at position 1. v a a) N S b) N E c) N W d) N N e) S E W N S E 2 1

Clicker Question 3.5.1 A car is moving counter-clockwise around a circular section of road at constant speed. What are the directions of its velocity and acceleration at position 1. v a a) N S b) N E c) N W d) N N e) S E W N S E 2! a! v 1

Clicker Question 3.5.2 A car is moving counter-clockwise around a circular section of road at constant speed. What are the directions of its velocity and acceleration at position 2? v a a) E S W N S E 1 b) E E c) E N 2 d) E W e) W S

Clicker Question 3.5.2 A car is moving counter-clockwise around a circular section of road at constant speed. What are the directions of its velocity and acceleration at position 2? v a a) E S b) E E c) E N W N S E! a 2! v 1 d) E W e) W S

Chapter 8 Accelerated Circular Motion

8.1 Rotational Motion and Angular Displacement Why are there 360 degrees in a circle? Why are there 60 minutes in an hour? Why are there 60 seconds in a minute? Because the Greeks, who invented these units were enamored with numbers that are divisible by most whole numbers, 12 or below (except 7 and 11). Strange, because later it was the Greeks who discovered that the ratio of the radius to the circumference of a circle was a number known as 2 π. A new unit, radians, is really useful for angles.

8.1 Rotational Motion and Angular Displacement A new unit, radians, is really useful for angles. Radian measure θ(radians) = s = rθ s (arc length) r (radius) (s in same units as r) θ s r Full circle θ = s r = 2π r r = 2π (radians) Conversion of degree to radian measure θ(rad) = θ(deg.) 2π 360 2π 360 rad deg. = 1 rad deg.

8.1 Rotational Motion and Angular Displacement Example: Adjacent Synchronous Satellites Synchronous satellites are put into an orbit whose radius is 4.23 10 7 m. If the angular separation of the two satellites is 2.00 degrees, find the arc length that separates them. Convert degree to radian measure 2.00deg Determine arc length 2π rad 360deg s = rθ = 4.23 10 7 m = 0.0349 rad ( ) 0.0349 rad = 1.48 10 6 m (920 miles) ( )

8.1 Rotational Motion and Angular Displacement For an observer on the earth, an eclipse can occur because angles subtended by the sun and the moon are the same. θ = s Sun r Sun s Moon r Moon 9.3 mrad

8.1 Rotational Motion and Angular Displacement The angle through which the object rotates is called the angular displacement vector Δθ = θ θ o SI unit of angular displacement, radian (rad) Simplified using θ o = 0, and Δθ = θ, angular displacement vector. r Δθ Vector Counter-clockwise is + displacement Clockwise is displacement θ 0 = 0

Clicker Question 8.1 Radian measure for angles Over the course of a day (twenty-four hours), what is the angular displacement of the second hand of a wrist watch in radians? a) 1440 rad b) 2880 rad c) 4520 rad d) 9050 rad e) 543,000 rad

Clicker Question 8.1 Radian measure for angles Over the course of a day (twenty-four hours), what is the angular displacement of the second hand of a wrist watch in radians? a) 1440 rad b) 2880 rad c) 4520 rad d) 9050 rad e) 543,000 rad Second hand makes 1 rev. in 1 minute. 1 day = (1 day)(60 min./hr)(24 hr/day) = 1440 min. # rev. = ( 1 rev./min. )(1440 min.) = 1440 rev. 1 rev. = 2π radians Total θ displacement = (2π rad/min.)(1440 min.) = 9048 rad

8.2 Angular Velocity and Angular Acceleration DEFINITION OF AVERAGE ANGULAR VELOCITY ω = Δθ Δt where Δt = t t o SI Unit of Angular Velocity: radian per second (rad/s) Δθ is the same at all radii. Δt is the same at all radii. ω = Δθ Δt is the same at all radii. r 2 r 1 Δθ in time Δt Δθ Angle change θ 0 = 0

8.2 Angular Velocity and Angular Acceleration ω Case 1: Constant angular velocity, ω. r ω = Δθ Δt Δθ = ω Δt Δθ θ 0 = 0 Example: A disk rotates with a constant angular velocity of +1 rad/s. What is the angular displacement of the disk in 13 seconds? How many rotations has the disk made in that time? Δθ = ω Δt = (+1 rad/s)(13 s) = +13 rad 2π radians = 1 rotation 2π rad/rot. n rot = Δθ 2π rad/rot. = 13 rad 6.3 rad/rot = 2.1 rot.

8.2 Angular Velocity and Angular Acceleration Case 2: Angular velocity, ω, changes in time. r ω Instantaneous angular velocity at time t. Δθ ω = lim Δt =0 Δt ω 0 DEFINITION OF AVERAGE ANGULAR ACCELERATION SI Unit of Angular acceleration: radian per second squared (rad/s 2 )

8.2 Angular Velocity and Angular Acceleration Example: A Jet Revving Its Engines As seen from the front of the engine, the fan blades are rotating with an angular speed of 110 rad/s. As the plane takes off, the angular velocity of the blades reaches 330 rad/s in a time of 14 s. Rotation is clockwise (negative) Find the angular acceleration, assuming it to be constant. ( ) ( 110rad s) α = 330rad s 14 s = 16rad s 2

8.2 The Equations of Rotational Kinematics The equations of rotational kinematics for constant angular acceleration: ANGULAR ACCELERATION ANGULAR VELOCITY ω = ω 0 + αt θ = 1 2 ( ω 0 + ω ) t TIME ANGULAR DISPLACEMENT ω 2 = ω 0 2 + 2αθ θ = ω 0 t + 1 2 αt 2

8.2 The Equations of Rotational Kinematics

8.2 The Equations of Rotational Kinematics Reasoning Strategy 1. Make a drawing. 2. Decide which directions are to be called positive (+) and negative ( ). 3. Write down the values that are given for any of the five kinematic variables. 4. Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation. 5. When the motion is divided into segments, remember that the final angular velocity of one segment is the initial angular velocity for the next. 6. Keep in mind that there may be two possible answers to a kinematics problem.

Clicker Question 8.2 Rotational motion kinematics Given the initial and final angular velocity of a disk, and the total angular displacement of the disk, with which single equation can the angular acceleration of the disk be obtained? a) ω = ω 0 + αt b) θ = 1 2 (ω + ω 0 )t c) θ = ω 0 t + 1 2 αt2 d) ω 2 = ω 0 2 + 2αθ e) none of the above

Clicker Question 8.2 Rotational motion kinematics Given the initial and final angular velocity of a disk and the total angular displacement of the disk, the angular acceleration of of the disk can be obtained with which single equation? a) ω = ω 0 + αt b) θ = 1 (ω + ω )t 2 0 c) θ = ω 0 t + 1 2 αt2 d) ω 2 = ω 2 0 + 2αθ e) none of the above No time given! ω 2 = ω 2 0 + 2αθ α = ω 2 2 ω 0 2θ

8.2 The Equations of Rotational Kinematics Example: A disk has an initial angular velocity of +375 rad/s. The disk accelerates and reaches a greater angular velocity after rotating ω through an angular displacement r of +44.0 rad. ω 0 If the angular acceleration has a constant value of +1740 rad/s 2, find the final angular velocity of the disk. Given: ω 0 = +375 rad/s,θ = +44 rad, α = 1740 rad/s 2 Want: final angular velocity, ω. No time! ω 2 = ω 0 2 + 2αθ = (375 rad/s) 2 + 2(1740 rad/s 2 )(+44 rad) ω = 542 rad/s