1. Rotational Kinematics Chapter 10 Herriman High AP Physics C

2. Section 10.2 The Rotational Variables θ (theta) – angular displacement – radians ω (omega) – angular velocity – radians/sec α (alpha) = angular acceleration – radians/sec 2 t – still time and still in seconds Herriman High AP Physics C

3. Measuring Angular Displacement θ = s/r 360° = 2 π radians = 1 revolution v= ωr = Linear velocity = angular velocity x radius a = αr Linear acceleration = angular acceleration x radius θ r s Herriman High AP Physics C

4. Sample Problems If an exploding fireworks shell makes a 10° angle in the sky and you know it is 2000 meters above your head, how many meters wide is the arc of the explosion? Herriman High AP Physics C

5. Solution θ = L/r θ = 10° x π rad/180° = 0.17 rad 0.17 rad = L/2000 meters L = 2000 meters x 0.17 rad = 349 meters Herriman High AP Physics C

6. Sample Problems Convert the following measures from Radians to degrees: 3.14 rad 150 rad 24 rad Herriman High AP Physics C

7. Solution 3.14 rad x 180°/ π rad = 180° 150 rad x 180°/ π rad = 8594° 24 rad x 180°/ π rad = 1375° Herriman High AP Physics C

8. Sample Problems What is the linear speed of a child seated 1.2 meters from the center of a merry-go-round if the ride makes one revolution in 4 seconds? What is the child’s acceleration? Herriman High AP Physics C

9. Solution a)v = ωr and ω=2π rad/4 sec = 1.6 rad/s since r = 1.2 m then v = 1.58 rad/s x 1.2 m = 1.9 m/s b) Since the linear velocity is not changing there is no linear or tangential acceleration, but the child is moving in a circle so there is centripetal or radial acceleration which you will recall fits the equation: a c = v 2 /r and since v = ωr a c = ω 2 r so a c = (1.9 m/s )2 /1.2 m = 3.0 m/s 2 or a c = ω 2 r = (1.6 rad/s) 2 (1.2 m) = 3.0 m/s 2 Herriman High AP Physics C

10. Section 10.3 - 10.5 Rotation with Constant Angular Acceleration & the Relationship Between Linear and Angular Acceleration LinearAngular V = v 0 +at ω=ω 0 +άt x = v 0 t+½at 2 θ =ω 0 t+½άt 2 V 2 = v 0 2 +2ax ω 2 =ω 0 2 +2άθ V avg = (v 0 +v f )/2 ω avg =(ω 0 +ω f )/2 Herriman High AP Physics C

11. Sample Problem A angular velocity of wheel changes from 10 rad/s to 30 rad/s in 5 seconds. a) What is its angular acceleration? b) What is its angular displacement while it is accelerating? Herriman High AP Physics C

12. Solution α = (ω f – ω 0 )/t = =(30 rad/s – 10 rad/s)/5 sec = 4 rad/s 2 θ = ω 0 t = ½αt 2 = 10 rad/s + ½(4 rad/s 2 )(5 sec) 2 = 60 rad Herriman High AP Physics C

13. Section 10.6 Kinetic Energy of Rotation K = ½ Iω 2 I = Rotational Inertia ω = Angular Velocity Rotational Inertia is a measure of how difficult it is to rotate an object on a given axis. This is determined by some ratio of the mass of the object and the radius of the object along the axis of rotation. Herriman High AP Physics C

14. Section 10.7 Calculating the Rotational Inertia I = ∫ r 2 dm For various objects the results of these calculations can be found on page 253 in table 10.2 Parallel axis Theorem If you know the I value for a body through the center of mass then a shortcut to finding its I value for a parallel axis is given as: I = I cm + Mh 2 Where h is the perpendicular distance between the axes. Herriman High AP Physics C

15. Alta High AP Physics Section 10.8 Torque Torque = Fr = force x radius Torque is measured in newtonmeters which means that it has the same units as work in a linear system. Herriman High AP Physics C

16. Alta High AP Physics Section 10.9 Newton’s Second Law for Rotation Torque = I α = rotational inertia x angular acceleration Hence Fr = I α Herriman High AP Physics C

17. Alta High AP Physics Sample Problem A force of 10 newtons is applied to the edge of a bicycle wheel (a thin ring mass 1 kg and radius of 0.5 meters). What is the resulting angular acceleration of the wheel? If the wheel was at rest when the force was applied and the force is applied for 0.4 seconds what is the angular velocity of the wheel immediately after it is applied? Herriman High AP Physics C

18. Alta High AP Physics Solution Since Fr = I α then α = Fr/I and since the wheel is a thin ring: I = mr 2 = (1 kg)(0.5 m) 2 = 0.25 kg m 2 So α = (10 N)(0.5 m)/0.25 kg m 2 = 20 rad/s 2 ω f = ω 0 + αt = 0 rad/s + (20 rad/s 2 )(0.4 sec) = 8 rad/s Herriman High AP Physics C

19. Section 10.10 Work and Rotational Kinetic Energy By Conservation of energy: ∆K = K f – K i = ½ Iω f 2 – ½ Iω i 2 = W W = ∫ r dθ W = Τ∆θ P = dθ/dt = Tω For a table of corresponding relationships Translational:Rotational see table 10.3 on page 261 Herriman High AP Physics C

20. Problems Types Finding angles when distance and length of arc are known Converting from revolutions to radians and radians to degrees Finding angular velocity and angular acceleration if radius and their linear counterparts are known Using angular kinematic equations Herriman High AP Physics C

21. Alta High AP Physics Problems Types Finding angles when distance and length of arc are known Converting from revolutions to radians and radians to degrees Finding angular velocity and angular acceleration if radius and their linear counterparts are known Using angular kinematic equations Calculating Torque Herriman High AP Physics C