1. Physics 111 Rotational Motion + inertia

2. Equations: Angular displacement ɵ = ɵ final - ɵ initial
ɵ = arc length (s) / r
2п= 1 full revolution
2пr = circumference
Angular velocity ᾠ = ɵ / t
Angular Acceleration ᾰ = ᾠ / t
Tangential Velocity (Vt) = r (ᾠ)
Tangential Acceleration (At) = r(ᾰ)
Centripetal Acceleration (Ac) = r (ᾠ^2)
Centripetal Force (Fc) = mr ᾠ^2
Impluse (I) = mr^2
Kinetic Energy = .5 I ᾠ^2
ᾠf = ᾠi+ ᾰ t
ɵ = .5 (ᾠi + ᾠf) t
Ɵf = ɵi + ᾠi(t) + .5 ᾰ t^2
ᾠf^2 = ᾠi^2+ 2ᾰ ɵ

3. Practice problem:
A motor slows from 34 rad/s to 10 rad/s while the turning the drum through an angle of 323 radians. What is the magnitude of the angular acceleration of the motor?

4. ωF^ 2 = ω0 ^2 + 2αΔθ 10^2 = 34^2 + 2(α)(323) Α= -1.63 rad/sec62
Solution:
ωF^ 2 = ω0 ^2 + 2αΔθ
10^2 = 34^2 + 2(α)(323)
Α= rad/sec62

5. More problems 
Passengers are seated in a horizontal circle of radius 4 m. The seats begin from rest and are uniformly accelerated for 13 seconds to a maximum rotational speed of .6 rad/s.
What is the tangential acceleration of the passengers during the first 30 s of the ride?

6. Solution: At = rᾰ ᾠf = ᾠi+ ᾰ t ᾰ = ᾠf /t =.6 / 13 = .046 rad/ sec ^2
At = rᾰ = 4* .046 = .185 m/sec^2
Vt = r ᾠ
ᾠ= .046 * 30 sec = 1.38 rad/s
Vt = r ᾠ = 4 * 1.38 = 5.52 m/s

7. Problem Part1:
A bicycle wheel of radius r = 1.5  m starts from rest and rolls 100 m without slipping in 30 s. Calculate
a) the number of revolutions the wheel makes
b) the number of radians through which it turns
c)The average angular velocity.

8. Solution:
a)If there is no slipping, the arc length through which a point of the rim moves is equal to the distance travelled, so that the number of revolutions is: n = 100 m / 2п(1.5) = 10.6
b) Delta ɵ= 2пn = s/r =100m/1.5 = 66.7  radians.
c)Average angular velocity: =  Delta ɵ / t = 66.7 / 30 * .5 seconds = 4.44  rads/s.

9. Part 2:
Assuming that the angular acceleration of the wheel given above was constant, calculate:
a) The angular acceleration
b) the final angular velocity
c) the tangential velocity and tangential acceleration of a point on the rim after one revolution.

10. Solution:
a)For constant angular acceleration:  Ɵf = ɵi + ᾠi(t) + .5 ᾰ t^2
Using  ᾠi= 0 and solving for  ᾰ gives: =  2(66.7)/(30^2) = .15 rad/ s^2
b) ᾠf = ᾠi+ ᾰ t  0 + (0.15)(30) = 4.5  rads/s
c)After one revolution,  delta Ɵ= 2 п  
ᾠf^2 = ᾠi^2+ 2ᾰ ɵ  1.37  rads/s.
The tangential velocity and acceleration are:
Vt= r ᾠ =(1.5  m )(1.37  rads/s ) = 2.06  m/s
at = r ᾰ = (1.5  m )(0.15 rads/s 2) = .225 m/s 2.