1. Aim: How do we explain rotational kinematics?

2. Angular distance vs. Translational distance
We can relate the angular distance traveled to the translational distance traveled by using S = rθ

3. Rotational Kinematics Equations
ωf=ωi + αt θf=θi +ωit +1/2αt2 ωf2=ωi2 +2α(θf-θi)

4. Relations between rotational and translational quantities
v=rω at=rα ac= v2/r = rω2

5. Thought Question 1
Consider again the pairs of angular positions for the rigid body listed in the previous thought questions. If the object starts from rest at the initial angular position, moves counterclockwise with constant angular acceleration, and arrives at the final angular position with the same angular speed, for which choice is the angular acceleration the highest?
a)3 rad, 6 rad b) -1 rad, 1 rad c) 1 rad, 5 rad
Use ωf2=ωi2+2αθ to analyze. Since each situation has the same values for ωi and ωf, we can see that the scenario with the smallest value of θ will have the highest angular acceleration. This is choice B) b

6. Rotating wheel problem
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed is 2.00 rad/s at t = 0.
Through what angles does the wheel rotate between t= 0s and t = 2s?
Θ=ωit+1/2αt2
Θ=2(2)+1/2(3.5)(2)2=11 rad
b) What is the angular speed of the wheel at t= 2s?
ωf=ωi+αt
ωf=2+(3.5)(2)= 9 rad/s
c)Find the angle through which the wheel rotates between t=2s and t=3s.
Θ=9(1)+1/2(3.5)(1)2=10.8 rad
11 rad 9 rad/s rad

7. Thought Question 2
When a wheel of radius R rotates about a fixed axis at a steady rate,
Do all the points on the wheel have the same angular speed? Yes
Do they all have the same tangential speed?No
If the angular speed is constant and equal to ω, describe the tangential speeds and total translational accelerations of the points located at r = 0, r=R/2, and r = R, where the points are measured from the center of wheel.
v=rω At r=0, v=0…At r=R/2, v=Rω/2…At r=R, v=Rω
ac=rω At r=0, ac=0…At r=R/2, ac=Rω2/2…At r=R, ac=Rω2
There is only centripetal (radial) acceleration and no tangential acceleration because there is no change in angular speed
Yes, no, v=rw/2 a=rw^2/ v=rw a=rw^2

8. Thought Question 3
A phonograph record is rotated so that the surface sweeps past the laser at a constant tangential speed. Consider two circular grooves of information on an LP-one near the outer edge and one near the inner edge. Suppose the outer groove “contains” 1.8s of music. Does the inner groove also contain the same time interval of music? The same amount of music is on the inner groove and outer groove but the spacing is different

9. Thought Question 4
The launch area for the European Space Agency is not in Europe-it is in South America. Why?
The angular speed of the Earth as it rotates is the same everywhere but the tangential speed of the Earth is fastest near the equator since it must spin through the greatest distance here. As a result, it is easiest to launch a rocket from an area near the equator since it already inherits a faster velocity from the ground itself.