7 Rotational Motion
© 2010 Pearson Education, Inc. Slide 7-2 © 2010 Pearson Education, Inc. Slide 7-3 Recall from Chapter 6… . Angular displacement = θ . θ= ω t . Angular Velocity = ω (Greek: Omega) . ω = 2 π f and ω = θ/ t . All points on a rotating object rotate through the same angle in the same time, and have the same frequency. . Angular velocity: all points on a rotating object have the same angular velocity, ω, but different speeds, v, and v =ωr. . v =ωr
© 2010 Pearson Education, Inc. . ω is positive if object is rotating counterclockwise. (Negative if rotation is clockwise.) . Conversion: 1 revolution = 2 π rad
© 2010 Pearson Education, Inc. Checking Understanding
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. A. The angular velocity of A is twice that of B. B. The angular velocity of A equals that of B. C. The angular velocity of A is half that of B.
© 2010 Pearson Education, Inc. Slide 7-13 Answer Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. A. The angular velocity of A is twice that of B. B. The angular velocity of A equals that of B. C. The angular velocity of A is half that of B. All points on the turntable rotate through the same angle in the same time. ω = θ/ t All points have the same period, therefore, all points have the same frequency. ω = 2 π f © 2010 Pearson Education, Inc. Slide 7-14 Checking Understanding
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. A. The speed of A is twice that of B. B. The speed of A equals that of B. C. The speed of A is half that of B.
© 2010 Pearson Education, Inc. Slide 7-15 Answer Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. A. The speed of A is twice that of B. B. The speed of A equals that of B. C. The speed, v, of A is half that of B. v = wr Twice the radius means twice the speed, v. Therefore, B moves with twice the speed of A.
© 2010 Pearson Education, Inc. Slide 7-16 Angular Acceleration, α (Greek: Alpha) Angular acceleration α measures how rapidly the angular velocity is changing: If the turning rate, or angular velocity, ω, is constant, then there is no angular acceleration, α, but if ω is changing, then the object is NOT undergoing UCM and the object has angular acceleration…
Units for Angular Acceleration are rad/s/s or rad/s2 © 2010 Pearson Education, Inc. Sign of the Angular Acceleration—careful!
-ω, +α = -ω , -α = +ω, -α = speeding up +ω, +α slowing down slowing down =speeding© 2010 Pearson Education, up Inc. Do NOT confuse angular acceleration, α = Δω/ Δt, with centripetal acceleration, 2 2 ac = v /r = ω r. . Angular acceleration, α, indicates how rapidly the angular velocity, ω, is changing. (No UCM)
. Centripetal acceleration, ac, is a vector quantity, directed toward the center of a particle’s circular path, and is nonzero even if the angular velocity (turning rate) is constant.
(ac exists always– whether experiencing UCM or not.) © 2010 Pearson Education, Inc. Linear and Circular Motion Compared
© 2010 Pearson Education, Inc. Slide 7-18 Linear and Circular Kinematics Compared
ωf = ωi +αt
© 2010 Pearson Education, Inc. Slide 7-19 Linear Motion Graphs related to Rotational Motion Graphs:
. Displacement vs. Time: Slope = Velocity
θ
t . Angular displacement, θ, vs. Time: Slope = Angular Velocity, ω= Δθ/Δt
© 2010 Pearson Education, Inc. Linear Motion Graphs related to Rotational Motion Graphs:
. Velocity vs. Time: Slope = Acceleration
ω
t . Angular velocity vs. Time, Slope = Angular Acceleration, α =Δω/Δt
© 2010 Pearson Education, Inc. Centripetal and Tangential Acceleration
No UCM:
© 2010 Pearson Education, Inc. Slide 7-22 Example Problem A high-speed drill rotating CCW takes 2.5 s to speed up to 2400 rpm.
A. What is the drill’s angular acceleration?
B. How many revolutions does it make as it reaches top speed?
© 2010 Pearson Education, Inc. Slide 7-21 Answer:
. Given: ωi = 0, ωf = 2400 rev/min, t = 2.5 seconds
. ωf = 2400 rev/min (2πrad/1 rev)(1 min/60 sec) = 251.3 rad/s
K j k . Find: α = ?, ; θ = # revolutions = ?
. Plan: Find α using ωf = ωi +αt, then find θusing 2 K j k ; θ=ωi t + 1/2 αt . Convert θ to revolutions.
. Solve: ωf = ωi +αt . 251.3 rad/s = 0 + α (2.5 s) . α= (251.3 rad/s)/(2.5 s) = 100.5 rad/s2 = α
. Go to next slide for more… © 2010 Pearson Education, Inc. 2 . Δθ=ωi t + 1/2 αt . Δθ= 0 + ½ (100.5 rad/s2)(2.5 s)2 . Δθ= 314.16 rad . Δθ = 314.16 rad (1 rev/2πrad) = 50 rev . Δθ= 50 revolutions
© 2010 Pearson Education, Inc. © 2010 Pearson Education, Inc. Torque… rhymes with fork… Every time you open a door, turn on a water faucet, or tighten a nut with a wrench, you exert a turning force. This turning force produces a torque. Torque is different from a force. If you want to make an object move, apply a force. Forces tend to make things accelerate. If you want to make an object turn or rotate, apply a torque. Torques produce rotation.
© 2010 Pearson Education, Inc. Interpreting Torque, ( (Greek: Tau) Torque is due to the component of the force perpendicular to the radial line.
t = rF^ = rFsinf © 2010 Pearson Education, Inc. Slide 7-25 A Second Interpretation of Torque
t = r^F = rFsinf © 2010 Pearson Education, Inc. Slide 7-26 Signs and Strengths of the Torque
© 2010 Pearson Education, Inc. Slide 7-27 Torque: that which causes rotational velocity to change. . Variable for torque= (tau) . Increasing F causes greater linear acceleration . Increasing causes greater rotational acceleration . Demonstration: pushing door open…When does door open best?
. A) from outside edge to axis of rotation . B) pushing toward hinge . C) at an angle . D) from center of door to axis of rotation (see next slide)
© 2010 Pearson Education, Inc. Checking Understanding The four forces shown have the same strength. Which force would be most effective in opening the door?
A. Force F1 B. Force F2 C. Force F3 D. Force F4 E. Either F1 or F3
© 2010 Pearson Education, Inc. Slide 7-23 Answer The four forces shown have the same strength. Which force would be most effective in opening the door?
A. Force F1 B. Force F2 C. Force F3 D. Force F4 E. Either F1 or F3
© 2010 Pearson Education, Inc. Slide 7-24 What does torque depend on?
1. The size of the force (how hard you push) 2. The point where the force is applied relative to the rotation point. 3. The direction of the force.
© 2010 Pearson Education, Inc. = F r Where F and r have to be to each other! . = F r or…
. = F r (Units: N·m) . F = the force applied . r = the moment arm or the lever arm = distance between axis of rotation and line of action so that where Torque is positive when the lever arm (r) and line of the force tends to produce action meet, a 90o angle is a counter-clockwise formed. rotation, and negative when force tends to . Pivot point = axis of rotation produce clockwise
©(in 2010 thisPearson case,Education, theInc. hinge) rotation. Even though the magnitude of the applied force is the same in each case, why does the bottom situation produce the greatest amount of torque?
© 2010 Pearson Education, Inc. For You to Solve:
. A person pushes on the edge of a 0.90m wide door with a force of 5.0N acting at an angle of 350 to the plane of the door. What is the torque of the force acting on the door?
© 2010 Pearson Education, Inc. © 2010 Pearson Education, Inc. Another problem for you to solve… Revolutionaries attempt to pull down a statue of the Great Leader by pulling on a rope tied to the top of his head. The statue is 17 m tall, and they pull with a force of 4200 N at an angle of 65° to the horizontal. What is the torque they exert on the statue? If they are standing to the right of the statue, is the torque positive or negative?
© 2010 Pearson Education, Inc. Slide 7-28 And the answer is…
. F = F (cosθ) F=? o . F = 4200N (cos 65 ) θ=65o
. F = 1775 N F = 4200N
. = F r . = (1775 N)(17 m) r = 17 m . = 30,175 Nm, clockwise . = - 30,175 Nm
© 2010 Pearson Education, Inc. Which factor does the torque on an object not depend on?
A. The magnitude of the applied force. B. The object’s angular velocity. C. The angle at which the force is applied. D. The distance from the axis to the point at which the force is applied.
© 2010 Pearson Education, Inc. Slide 7-7 Answer Which factor does the torque on an object not depend on?
A. The magnitude of the applied force. B.The object’s angular velocity. C. The angle at which the force is applied. D. The distance from the axis to the point at which the force is applied.
© 2010 Pearson Education, Inc. Slide 7-8 © 2010 Pearson Education, Inc. © 2010 Pearson Education, Inc. Slide 7-4 A net torque applied to an object causes…??? A. a linear acceleration of the object. B. the object to rotate at a constant rate. C. the angular velocity of the object to change. D. the moment of inertia of the object to change.
© 2010 Pearson Education, Inc. Slide 7-11 Answer A net torque applied to an object causes… A. a linear acceleration of the object. B. the object to rotate at a constant rate. C.the angular velocity of the object to change. D. the moment of inertia of the object to change.
© 2010 Pearson Education, Inc. Slide 7-12