Aim: How do we explain rotational kinematics?

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

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

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

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

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

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/2 d=20.1 m a)126 rad/s b) 3.77 m/s c)1.26 km/s2 d)20.1m

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