
Possibly useful information:
s = s_{i} + v _{i} t + (^{1}/_{2}) a t^{2}  a_{c} = v ^{2} / r  p = m v 
v = v _{i} + a t  W = F s cos  P_{Tot} = p_{1} + p_{2} 
v ^{2} = v_{i }^{2} + 2 a (s  s_{i})  W = Fs  P_{Tot,i} = P_{Tot,f} 
sin = opp/hyp  AB = A B cos  
cos = adj/hyp  AB = A_{x} B _{x} + A_{y} B _{y} + A_{z} B_{z}  
tan = opp/adj  dW = F _{x} dx 
M = m_{i } 
F = m a  K = (1/2) m v^{2}  
F_{12} =  F_{21}  W_{net} = K  
f_{smax} = µs F_{n}  P = W/t  
f_{k} = µ_{k} F _{n}  U =  W  
g = 9.8 m/s ^{2} Å 10 m/s ^{2}  E = K + U  
E_{i }= E_{f}  
F_{x }=  [ dU / dx ]  
dU =  F _{x }dx  
F_{g} = G Mm / r^{2} 
1. Momentum
Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in a perfectly elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.00 m/s. After the collision, the orange disk moves along a direction that makes an angle of 37.0° with its initial direction of motion, and the velocity of the yellow disk is perpendicular to that of the orange disk (after the collision  and this will always be the case for elastic collisions between equal masses). Determine the final speed of each disk.
This is problem 9.45.
We can substitute 5.0 m/s = v_{o} at the end.
P_{fx} = m v_{org,x} + m v_{yel,x}
P_{fx} = m v_{org} cos 37^{o} + m v_{yel} cos 53^{o}
P_{fx} = m v_{org} (0.8) + m v_{yel} (0.6)
P_{fx} = m [ v_{org} (0.8) + v_{yel} (0.6) ]
P_{fx} = m [ v_{org} (0.8) + v_{yel} (0.6) ] = m v_{o} = P_{ix}
v_{org} (0.8) + v_{yel} (0.6) = v_{o}
Now, for the ycomponent,
P_{fy} = m v_{org,y} + m v_{yel,y}
P_{fy} = m v_{org} sin 37^{o}  m v_{yel} sin 53^{o}
P_{fy} = m v_{org} (0.6)  m v_{yel} (0.8)
P_{fy} = m [ v_{org} (0.6)  v_{yel} (0.6) ]
P_{fy} = m [ v_{org} (0.6)  v_{yel} (0.8) ] = 0 = P_{iy}
[ v_{org} (0.6)  v_{yel} (0.8) ] = 0
Now we have two equations and two unknowns, v_{org} and v_{yel}.
0.6 v_{org} = 0.8 v_{yel}
v_{org} = 1.333 v_{yel}
v_{org} (0.8) + v_{yel} (0.6) = v_{o}
(0.8)(1.33 v_{yel }) + 0.6 v_{yel} = v _{o}
1.667 v_{yel} = v _{o}
v_{yel} = 0.6 v _{o}
v_{yel} = 0.6 (5.0 m/s) = 3.0 m/s
v_{org} = (1.333) (3.0 m/s)
v_{yel} = 4.0 m/s
2. Rotation
A 15kg mass and a 10kg mass are suspended by a pulley that has a radius of 10 cm and a mass of 3.0 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 3.0 m apart. Treat the pulley as a uniform disk.
I_{disk }= (^{1}/_{2}) M R^{2}
Detemine the speeds of the two masses as they pass each other.
This is problem 10.32
Now, draw freebody diagrams for all three bodies,
We will need the moment of inertia of the pulley,
I = (1/2) M r^{2} = (0.5) (3.0 kg) (0.10 m)^{2} = 0.015 kg m^{2}
For mass m_{2}, I have taken up to be positive; that means
F_{net,2} = T_{2}  m_{2} g = m_{2} a
For the pulley,
_{net} = _{2}  _{1} = I
_{1} = r T_{1} = (0.10 m) T_{1}
_{2} = r T_{2} = (0.10 m) T_{2}
_{net} = _{1}  _{2} = (0.10 m) T_{1}  (0.10 m) T_{2} = I
(0.10 m) [ T_{1}  T_{2} ] = I = (0.015 kg m^{2})
The linear acceleration a and the angular acceleration are related by
a = r
or
= a / r
so we may write
(0.10 m) [ T_{1}  T_{2} ] = (0.015 kg m^{2}) a / 0.10 m
(0.10 m) [ T_{1}  T_{2} ] = (0.15 kg m) a
T_{1}  T_{2} = (1.5 kg) a
And, for mass m_{1}, where I have taken down as positive,
F_{Net,1} = m_{1} g  T_{1} = m_{1} a
(15.0 kg) (9.8 m/s^{2})  T_{1} = (15.0 kg) a
147 N  T_{1} = (15.0 kg) a
Now we have three equations  from the three freebody diagrams  and three unknowns.
T_{1} = 147 N  (15.0 kg) a
T_{1}  T_{2} = (1.5 kg) a
[ 147 N  (15.0 kg) a]  T_{2} = (1.5 kg) a
T_{2} = [ 147 N  (15.0 kg) a]  (1.5 kg) a
T_{2} = 147 N  (16.5 kg) a
T_{2}  m_{2} g = m_{2} a
[147 N  (16.5 kg) a]  (10.0 kg) (9.8 m/s^{2}) = (10.0 kg) a
147 N  98 N = (26.5 kg) a
a = 49 N / 26.5 kg
a = 1.85 m/s^{2}
With this acceleration of 1.85 m/s^{2}, how fast will the blocks be moving after they have moved a distance of 1.5 m?
v^{2} = v_{o}^{2} + 2 a (y  y_{o})
v^{2} = 0^{2} + 2 (1.85 m/s^{2}) (1.5 m)
v^{2} = 5.55 m^{2} / s^{2}
v = 2.36 m/s
Alternate method, using Energy Conservation
E_{i} = E_{f}
E_{i} = U_{1i} = (15 kg) (9.8 m/s^{2}) (3.0 m) = 441 J
This takes the reference position to be the "ground", where the 10 kg mass is sitting. Since everything starts at rest, there is no kinetic energy at all.
E_{f} = K_{1} + U_{1} + K_{2} + U_{2} + K_{3}
K_{3} = K_{rot} = (^{1}/_{2}) I ^{2}
I = = (^{1}/_{2}) M R^{2} = (0.5) (3.0 kg) (0.10 m)^{2} = 0.015 kg m^{2}
K_{3} = K_{rot} = (^{1}/_{2}) I ^{2} = (0.5) (0.015 kg m^{2}) ^{2} = (0.0075 kg m^{2}) ^{2}
K_{2} = (^{1}/_{2}) m_{2} v^{2} = (0.5) (10 kg) v^{2} = (5 kg) v^{2}
K_{1} = (^{1}/_{2}) m_{1} v^{2} = (0.5) (15 kg) v^{2} = (7.5 kg) v^{2}
These v's are, indeed, the same. And these v's  this velocity  are/is related to the angular velocity by
v = R = (0.10 m)
= v / R = v / 0.10 m
K_{3} = K_{rot} = (0.0075 kg m^{2}) ( v / 0.10 m )^{2} = 0.75 kg v^{2}
U_{2} = m_{2} g h_{2} = (10 kg) (9.8 m/s^{2}) (1.5 m) = 147 J
U_{1} = m_{1} g h_{1} = (15 kg) (9.8 m/s^{2}) (1.5 m) = 220.5 J
E_{f} = K_{1} + U_{1} + K_{2} + U_{2} + K_{3}
E_{f} = 7.5 kg v^{2} + 220.5 J + 5 kg v^{2} + 147 J + 0.75 kg v^{2}
E_{f} = 13.25 kg v^{2} + 367.5 J
E_{f} = E_{i}
13.25 kg v^{2} + 367.5 J = 441 J
13.25 kg v^{2} = 73.5 J
v^{2} = 5.55 m^{2}/s^{2}
v = 2.36 m/s
This is an example of the "joy of mathematics"  mathematics works because it really does describe the "real world". This is the same answer we got earlier  by a totally different method! The ideas of Physics and mathematics are useful  and, therefore, important and worthwhile  because they explain and predict the "real world"!
3. Angular Momentum
A merrygoround of radius R = 2.0 m has a moment of inertia I = 250 kg m2 and is rotating at 10 rev/min. A 25kg child jumps onto the edge of the merrygoround.
What is the new angular speed of the merrygoround?
This is problem 11.28
L_{f} = L_{i}
L _{i} = I _{i} _{i}
I_{i} = 250 kg m^{2}
_{i} = 10 rev/min
L _{i} = (250 kg m^{2}) ( 10 rev/min) = 2 500 (kg m^{2} rev / min)
L _{f} = I _{f} _{f}
I_{f}= I_{MgR} + I_{child}
I_{child} = m r^{2} = (25 kg) (2.0 m)^{2} = 100 kg m^{2}
I _{f} = (250 + 100) kg m^{2} = 350 kg m^{2}
_{f} = L _{f} / I _{f} = L _{i} / I _{f} = [ 2 500 kg m^{2} rev / min ] / [ 350 kg m^{2} ]
_{f} = 7.14 rev / min
4. Calculus application
Show, by direct calculation, that the moment of inertia for a uniform rod, rotated about its Center of Mass, is (^{1}/_{12}) M L^{2}.
This one is "new". This is not a duplicate of a homework problem but is quite similar to ones we did in class.
I =r^{2} dm = x^{2} dx = x^{2} dx
I = x^{2} dx = (^{1}/_{3}) x^{3}
I = (^{1}/_{3}) [ ( L / 2 )^{3}  ( L / 2 )^{3} ] = (^{1}/_{3}) [ L^{3} / 4 ]
= M / L
I = [ M / L ] (^{1}/_{3}) [ L^{3} / 4 ]
I = (^{1}/_{12}) M L^{2}
(C) 1997, Doug Davis; all rights reserved