1. Aim: How do we explain rotational kinematics?

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

3. 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] The answer is b. Looking at the equation: ωf2=ωi2 +2α(θf-θi). Since each rotation has the same final angular velocity and initial angular velocity, we see that the rotation which has the smallest change in angle, must have the greatest angular acceleration. b

4. 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=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)=9.5 rad/s
c)Find the angle through which the wheel rotates between t=2s and t=3s.
Θ=ωit+1/2αt2 =2(3)+1/2(3.5)(3)2=21.75 rad
So between 2 and 3 seconds, θ=21.75 rad-11 rad=10.75 rad
11 rad 9 rad/s rad

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

6. Thought Question 2
When a wheel of radius R rotates about a fixed axis,
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.
At r=0, v=rω=0(0)= ac=rω2=r(0)2=0
At r=R/2, v=rω=(R/2)ω ac=(R/2)ω2
At r=R, v=rω=Rω ac=Rω2
Yes, no, v=rw/2 a=rw^2/ v=rw a=rw^2

7. 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 must be contained in each groove.

8. Thought Question 4
The launch area for the European Space Agency is not in Europe-it is in South America. Why? Near the equator in South America, the Earth is spinning faster than anywhere in Europe. Therefore, a space craft does not need to be accelerated to as great a speed from the equator as from Europe.

9. Problem 2
A disk 8.00 cm in radius rotates about its central axis at a constant rate of 1200 rev/min. Determine
Its angular speed
Its tangential speed at a point 3.00 cm from its center
The radial acceleration of a point on the rim
The total distance a point on the rim moves in 2.00s
ω=1200(2π/60)= 40π rad/s or m/s
v=rω=0.03(125.6)=3.768 m/s
ac=rω2=0.08(125.6)2=1,262 m/s2
ω=θ/t =θ/ so θ= rad
d=rθ d=0.08(151.2)= m
a)126 rad/s b) 3.77 m/s c)1.26 km/s2 d)20.1m

10. Problem 3
A car accelerates uniformly from rest and reaches a speed of 22 m/s in 9s. If the diameter of the tire is 58.0 cm, find
The number of revolutions the tires makes during this motion, assuming that no slipping occurs.
What is the final rotational speed of a tire in revolutions per second?
a) r=29 cm=0.29 m
a=Δv/Δt=22/9=2.4 m/s2
a=rα
2.4=0.29α α=8.3 rad/s2
Θ=ωit+1/2αt2
Θ=0(9)+1/2(8.3)92= rad
We can divide this by 2π which is the number of radians in one revolution to find the number of revolutions /(2π)=53.52 revolutions
b) v=rω
22=0.29ω
ω=75.86 rad/s …To convert into revolutions per second, divide by 2π, and ω=12 rev/s
a) 54.3 rev b) 12.1 rev/s