1. Aim: How do we explain rotational kinematics?

2. Are angular quantities vectors?
Yes, but there are only two directions that are possible: clockwise and counterclockwise

3. Which object moves with a constant nonzero angular acceleration?
ω=3
ω=4t2 +2t-6
ω=3t-4
ω=5t2 -4
C is the answer

4. Distances
How does the distance the center of mass of a car moves compare to the distance that the wheel travels? SAME

5. Thought Question 1
The graph below shows how the angular velocity of a rotating disk varies with time.
Rank the tangential accelerations of a point on the rim for moments a,b,c,and d. (Greatest to Least) (Assume positive values are greater than zero which are greater than negative values) a,b=c,d…Look at slope to find angular acceleration. Tangential acceleration is proportional to angular acceleration
b) Rank the radial accelerations of a point on the rim for moments a,b,c,and d. (Greatest to least)
b,a=c,d Radial acceleration only depends on the value of ω

6. Thought Question 2
A cockroach is on a rotating platform (like a merry go round). It spins in a counterclockwise direction with decreasing speed. At this moment,
What is the direction of the cockroach’s tangential acceleration? UP
What is the direction of the cockroach’s radial acceleration?
Right

7. Problem 1
A disk 8.00 cm in radius rotates about its central axis at a constant rate of 1200 rev/min. Determine
Its angular speed
ω=1200rev/min (2π/60)= 126 rad/s (There are 2π radians per revolution and 60 seconds per minute)
b) Its tangential speed at a point 3.00 cm from its center
v=rω=(0.03 m)(126)= 3.77 m/s
c)The radial acceleration of a point on the rim
ac =rω2=(0.08)(126)2=126,000 m/s2
d) The total distance a point on the rim moves in 2.00s
v=rω=d/t
(0.08)(126)=d/ d=20.1 m
a)126 rad/s b) 3.77 m/s c)1.26 km/s2 d)20.1m

8. Problem 2
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.
The radius of a tire is 0.29 m. The circumference of the tire is C=2(3.14)(.29)
The acceleration of the car is a=Δv/t=22/9=2.44 m/s2
d=vit+1/2at2=0(9)+1/2(2.44)(9)2=99 m
If we divide the distance traveled by the circumference around a tire, we get the number of revolutions which is about 54.3
b) What is the final rotational speed of a tire in revolutions per second?
vf=rωf
22=(0.29)ωf so ωf=75.86 rad/s and we can convert this into revolutions per seconds by dividing this by 2π since this is the number of radians per revolution and we get 12.1 rev/s as our answer
a) 54.3 rev b) 12.1 rev/s