1. Rotational Dynamics

2. Consider a bicycle wheel to be a ring of radius 30 cm and mass 1. 5 kg
Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 20 N is applied tangentially to a sprocket of radius 4.0 cm for 4.0 s, what linear speed does the wheel achieve, assuming it rolls without slipping?
a. 3.0 m/s
b. 5.9 m/s
c. 7.1 m/s
d. 24 m/s

3. M = 1.5 kg, R = 0.30 m, F = 20 N, r = 0.04 m
t = 4 s
Thin hoop, I = MR2 = kg m2
t = Ia
= t/I
= 20 N * 0.04 m/0.135 kg m2
= 5.9 rad/s2
= at = 5.9 * 4 = 23.6 rad/s
v = wr = rad/s * 0.3 m = 7.1 m/s

4. A wheel of moment of inertia of 5
A wheel of moment of inertia of 5.00 kgm2 starts from rest and accelerates under a constant torque of 3.00 Nm for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s?
a J
b J
c J
d. 122 J

5. I = 5 kgm2, t = 3.00 Nm , t = 8.00 s
Krot = ???
Krot = ½ I w2 = 1/2(5)w2
To find omega, w, use w = wo + at.
How do we find alpha, a?
a = t/I= 3/5 = 0.6 rad/s2.
So, w = wo + at = * 8 = 4.8 rad/s
So,
Krot = ½ I w2 = 1/2(5)w2 = 57.6 J

6. A solid sphere of mass 1. 0 kg and radius 0
A solid sphere of mass 1.0 kg and radius m is released from the top of a 1.0-m high 370 inclined plane. What is the speed of the sphere when it reaches the bottom of the inclined plane?
a. 3.7 m/s
b. 4.4 m/s
c. 5.6 m/s
d. 6.3 m/s

7. M =1.0 kg, R=0.010 m, y = 1.0 m
V = ???
First, let’s set up the linear and rotational dynamics equations for this sphere.
F = Ma
Mgsinq – f = Ma
Mgsinq –Ma = f
= Ia
rf = I a/r
f = Ia/r2 = 2/5MR2 (a/R2) = 2/5 Ma
Combining, Mgsinq – Ma = f= 2/5 Ma
Gives a = 5/7 g sinq = 4.2 m/s2

8. v = vo + at = m/s2 * t.
Find t from, x = ½ a t2 ,
t= (2x/a)1/2 = (2*1.66 m/ 4.2 m/s2)1/2 = 0.89 s
So, v = 3.7 m/s.

9. Rotational Dynamics Torque, τ =I α Work, W = τΘ
Kinetic Energy of Rotation, K = ½ Iω2
Conservation of Energy
Angular Momentum, L = Iω
Conservation of Angular Momentum

10. Newton’s 2nd Law
Linear motion
Fnet = ma
Linear motion
τnet = Iα

11. Fnet = m1 a Fnet = T1 - µN N = m1g T1 - µm1g=m1a T1 = m1a -µm1g
τnet = Iα
τnet =R(T2 – T1)
I = ½ MR2
R(T2 – T1) =½MR2α N
Accel. = a
Ang. Accel = α
f= µN T1
M,R T2
Fnet = m2 a
Fnet = m2g – T2
m2g –T2 =m2a
T2 = -m2a +m2g
Accel = a
m1g
m2g

12. -(m2 + m1)a +g(m2 +µm1) =½MRa/R
T1= m1a -µm1g;T2 – T1 =½MRα;T2= -m2a +m2g
-m2a +m2g – m1a +µm1g =½MRα
-(m2 + m1)a +g(m2 +µm1) =½MRa/R
(m2 + m1 +½M)a = g(µm1+ m2)
a = g(µm1+ m2)/(m2 + m1 +½M) N
Accel. = a
Ang. Accel = α
f= µN T1
M,R T2
Accel = a
m1g
m2g

13. What if we ignore pulley? a = 3.92 m/s2, 6.7% higher!
a = g(µm1+ m2)/(m2 + m1 +½M)
Let M =0.2 kg, m1=1 kg, m2=0.5 kg, µ=0.1
a = 3.7 m/s2,
What if we ignore pulley?
a = m/s2, 6.7% higher! N
Accel. = a
Ang. Accel = α
f= µN T1
M,R T2
Accel = a
m1g
m2g

14. It took work to rotate the pulley
The pulley goes from zero angular speed to non-zero angular speed, ω = αt.
This means that the pulley’s final kinetic energy, Kf , is ½ Iω2
Kf = ½(½ MR2)ω2
The work-energy theorem, W = ΔK
W = τΘ=¼MR2ω2

15. Torque and Angular Momentum
Recall, F = ma = mΔv/Δt = Δp/Δt, so making the correspondence to rotational motion,
τ = Iα = IΔω/Δt = Δ(Iω)/Δt = ΔL/Δt, where
L = Iω = angular momentum.
τ = ΔL/Δt
When no external forces are acting,
torque = 0, so
ΔL/Δt = 0 which means, Lf = Lo or
Ifωf = Ioωo