PHYS 1101 Collisions Solutions

Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Physics 1101

Collisions

Linear Collisions

The diagrams below are graphs of Force in kiloNewtons versus time in milliseconds for the motion of a 5-kg block moving to the right at 4.0 m/s.
(a) What is the magnitude and direction of the impulse acting on the block in each case?
(b) What is the magnitude and direction of the average force acting on the block in each case?
(c) What is the magnitude and direction of the final velocity of the block in each case?
1. Impulse is given by the area under the F-t curves. Since we have simple shapes, it is easy to find the area. For rectangles area is height × base and for triangles area is half the height × base.
  1. I = 3 kN × 3 ms = 9 N-s
  2. I = −1 kN × 6 ms = −6 N-s
  3. I = 2 kN × 2 ms + ½(−4 kN) × 2 ms = 0 N-s
  4. I = ½(4 kN) × 4 ms = 8 N-s
  If the impulse is positive, the net area was above the curve and it is directed to the right, if negative to the left.
2. We know I = F_aveΔt where Δt is how long the collision lasts. We read Δt from the graphs, so F_ave = I/Δt.
  1. F_ave = (9 N-s)/(3 ms) = 3000 N
  2. F_ave = (−6 N-s)/(6 ms) = -1000 N
  3. F_ave = (0 N-s)/(4 ms) = 0 N
  4. F_ave = (8 N-s)/(4 ms) = 2000 N
  If the average force is positive it is directed to the right, if negative to the left. The impulse and force have the same direction.
3. Impulse is also equal to the difference in momentum, I = mv_f − mv_i. We can rearrange our equation for v_f, v_f = I/m + v_i.
  1. v_f = (9 N-s)/(5.0 kg) + 4 m/s = 5.8 m/s
  2. v_f = (?6 N-s)/(5.0 kg) + 4 m/s = 2.8 m/s
  3. v_f = (0 N-s)/(5.0 kg) + 4 m/s = 4 m/s
  4. v_f = (8 N-s)/(5.0 kg) + 4 m/s = 5.6 m/s

The diagrams below are the velocity versus time graphs for the collision of motion of a 4-kg block with a wall. The collision lasts for 20 milliseconds in each case.
(a) What is the magnitude and direction of the impulse acting on the block in each case?
(b) What is the magnitude and direction of the average force acting on the block in each case?
1. Impulse is also equal to the difference in momentum, I = mv_f − mv_i. We have the mass, m = 4 kg.
  1. I = (4 kg) × (−6 m/s − 6 m/s) = −48 N-s
  2. I = (4 kg) × (2 m/s − 8 m/s) = −24 N-s
  3. I = (4 kg) × (6 m/s − 0 m/s) = +24 N-s
  If the impulse is positive it is directed to the right, if negative to the left.
2. We know I = F_aveΔt where Δt is how long the collision lasts. We have already calculated I and we are given Δt = 20 ms, so F_ave = I/Δt.
  1. F_ave = (−48 N-s)/(20 ms) = −2400 N
  2. F_ave = (−24 N-s)/(20 ms) = −1200 N
  3. F_ave = (+24 N-s)/(20 ms) = +1200 N

You've been rowdy and obnoxious in a bar and are now in the process of being thrown out by the scruff of the neck by the bouncer. The bouncer has hold of you for 5.0 s and you are take from a seated position to a final speed of 2.75 m/s. If your mass is 70.0 kg, what was your final momentum? What impulse and average force did the bouncer exert on your person? Assume all motion is in a straight line.
Momentum is defined by p = mv. Taking the direction of motion as positive, your initial momentum was zero and your final momentum is

p = (70.0 kg)(2.75 m/s) = 192.5 kg-m/s .
Impulse is defined as the change in momentum

I = p_f - p_i = 192.5 kg-m/s .
Average force is related to impulse by I = F_averageΔt, so

F_average = I / Δt = 192.5 kg-m/s / 5 s = 38.5 N .
This is the average force exerted on you and is in the same direction as your motion.
A ball of mass 0.500 kg with speed 15.0 m/s collides with a wall and bounces back with a speed of 10.5 m/s. If the motion is in a straight line, calculate the initial and final momenta and impulse. If the ball exerted an average force of 1000 N on the wall, how long did the collision last?

Momentum is defined by p = mv. Taking the right as positive, the initial momentum of the ball is

p_i = (0.5 kg)(-15 m/s) = -7.5 kg-m/s .
The final momentum is

p_f = (0.5 kg)(10.5 m/s) = 5.25 kg-m/s .
Impulse is defined as the change in momentum

I = p_f - p_i = 12.75 kg-m/s .

Average force is related to impulse by I = F_averageΔt, and the wall would exert this force on the ball to the right. Therefore

Δt = I / F_average = 12.75 kg-m/s / +1000 = 0.013 s.

The ball is in contact with the wall for approximately 13 milliseconds.
A ball of mass 0.25 kg glances of a wall as shown in the diagram. The ball approaches at 15 m/s at θ = 30° and leaves at 12 m/s at φ = 20°. The collision lasts for 15 milliseconds.
(a) What are the components of the impulse experienced by the ball?
(b) What are the components of the average force acting on the ball?
1. We know Impulse is equal to the difference in momentum, I = mv_f − v_i. This is a vector equation and to get components we consider the x and y components separately.
  
  I_x = mv_fx − mv_ix = (0.25) × (12cos20° − 15cos30°) = −0.4285 N-s
  
  I_y = mv_fy − mv_iy = (0.25) × (12sin20° − (−15sin30°)) = +2.9011 N-s
  
  or I = −i0.4285 + j2.9011 N-s.
2. We know I = F_aveΔt where Δt is how long the collision lasts. We have already calculated I and we are given Δt = 15 ms, so F_ave = I/Δt
  
  so F_ave = (−i0.4285 + j2.9011 N-s) / (15 ms) = −i28.6 + j193.4 N.

Explain why a person wearing a seatbelt in a car accident is less likely to be seriously hurt than the person who isn't wearing a seatbelt.
Wearing a seatbelt would not effect the person's initial and final momentum. Since impulse is the change in momentum, the impulse experienced by the person would be the same in either case. However, it is not impulse which is dangerous but the magnitude of the forces acting on the person's body. Impulse and average force are related by I = F_average Δt. For the same impulse, the average force will be higher as the collision time Δt decreases. Cars are designed to crumple on impact. This crumpling is designed to make a collision last as long as possible. If a person is wearing a seatbelt, the time for the person's change in momentum is the same as that of the car thereby minimizing the force on the occupant. If the person is not wearing a seatbelt, the Law of Inertia dictates that the person will keep moving forward (Note some people describe this by saying that the person was thrown forward by the force of impact. Why is this wrong?). This means that the person will impact the steering wheel or windshield. The steering wheel and windshield can't collapse as nicely as the front of the car so Δt is much smaller and thus the average force is much higher.
A lion of mass 120 kg leaps at a hunter with a horizontal velocity of 12m/s. The hunter has an automatic rifle firing bullets of mass 15 g with a muzzle speed of 630m/s and he attempts to stop the lion in midair. How many bullets would the hunter have to fire into the lion to stop its horizontal motion? Assume the bullets stick inside the lion.
The total momentum of the system, the bullets and the lion, would be zero since there are no external forces to consider,

P = m_lionv_lion - nm_bulletv_bullet = 0 ,

where the lion is assumed to be moving in the positive direction. Rearranging the equation to find n,

n = m_lionv_lion / m_bulletv_bullet = (120 kg 12 m/s) / (0.015kg 630 m/s) = 152.4 .

It would take 153 bullets to stop the lion dead, so to speak.
On a frictionless surface, a 6.0-kg rock approaches from the left at 3.5 m/s. It collides elastically with a 9.0-kg rock which is approaching from the right at 1.7 m/s. Find the final velocities of the rocks.
We are dealing with a collision, so we know we must conserve momentum,

m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i . (1)
We are told that the collision is elastic which means that kinetic energy is conserved. For a 1D collision, this is the same as

v_2f - v_1f = -(v_2i - v_1i) . (2)
Substituting in the give data, our equations become

6v_1f + 9v_2f = 6×3.5 + 9×(-1.7) = 5.7 ,

v_2f - v_1f = -(-1.7 - 3.5) = 5.2 .
Solving the two equations in two unknowns, we find v_1f = -2.74 m/s and v_2f = 2.46 m/s.
If the collision in question #8 had been perfectly inelastic, what would have been the final velocity of the rocks? How much kinetic energy would have been lost in the collision?
In any kind of collision, momentum is conserved so

m₁v_1f + m₂v_2f = m₁v_1i + m₂v_2i .
We are told that the collision is perfectly elastic which means that kinetic energy is not conserved and that the rocks stick together, v_1f = v_2f = v_f. Our equation becomes

(m₁ + m₂)v_f = m₁v_1i + m₂v_2i .
This allows us to find v_f immediately

v_f = [m₁v_1i + m₂v_2i]/(m₁ + m₂) = [6 kg×3.5 m/s + 9 kg×(-1.7 m/s)]/ 15 kg = 0.38 m/s .
The initial kinetic energy is

K_i = ½m₁(v_1i)² + ½m₂(v_2i)² = ½(6)(3.5)² + ½(9)(-1.7)² = 49.755 J .
The final energy is

K_f = ½(m₁ + m₂)(v_f)² = ½(6+9)(0.38)²= 1.083 J .
The difference in energy is

ΔK = K_f - K_i = -48.7 J .
A 50.0-kg skater is traveling due east at 3.00 m/s. A 70.0-kg skater is moving due south at 7.00 m/s. They collide and hold on to one another after the collision. Determine the magnitude and direction of their velocity after the collision. Ignore the effects of friction.

In any kind of free collision, momentum is conserved so

(m₁ + m₂)v_f = m₁v_1i + m₂v_2i .           (1)
The vector diagram for this equation is

Now momentum and velocity are vector quantities and the i and j components must be handled separately

(m₁ + m₂)v_fx = m₁v_1ix + m₂v_2ix ,          (1a)

(m₁ + m₂)v_fy = m₁v_1iy + m₂v_2iy .           (1b)
So we can rearrange these equations to find the components of the final velocity

v_fx = (m₁v_1ix + m₂v_2ix) / (m₁ + m₂) ,          (2a)

v_fy = (m₁v_1iy + m₂v_2iy) / (m₁ + m₂) .           (2b)
Using the given values, we find

v_fx = [(50 kg)(3 m/s) + (70 kg)(0)] / (50 kg + 70kg) = 1.25 m/s ,           (2a)

v_fy = [(50 kg)(0) + (70 kg)(-7 m/s) / (50 kg + 70 kg) = 4.083 m/s .         (2b)
To find the magnitude and direction of the final velocity, we use Pythagoras' Theorem and trigonometry,

v_f = [(v_fx)² + (v_fx)²]^½ = 4.27 m/s, and

θ = tan^-1 (|v_fy/v_fx|) = 72.98°.
The final velocity of the pair is 4.27 m/s at 73.0° south of east.
Football player A tackles and holds onto player B in the diagram below.
1. Assuming friction is negligible, what is the velocity of the players just after the collision?
2. If the collision lasts for 0.30 seconds, what average force (magnitude and direction) does player A exert on player B?
In any kind of free collision, momentum is conserved so

(m_A + m_B)v_f = m_Av_Ai + m_Bv_Bi .           (1)
The vector diagram for this problem is

Now momentum and velocity are vector quantities and the i and j components must be handled separately

(m_A + m_B)v_fx = m_Av_Aix + m_Bv_Bix ,          (1a)

(m_A + m_B)v_fy = m_Av_Aiy + m_Bv_Biy .           (1b)
So we can rearrange these equations to find the components of the final velocity

v_fx = (m_Av_Aix + m_Bv_Bix) / (m_A + m_B) ,          (2a)

v_fy = (m_Av_Aiy + m_Bv_Biy) / (m_A + m_B) .           (2b)
Using the given values, we find

v_fx = [(75 kg)(2.75 m/s)cos(50°) + (70 kg)(2.30 m/s)] / (75 kg + 70kg) = 2.025 m/s ,           (2a)

v_fy = [(75 kg)(2.75 m/s)sin(50°) + (70 kg)(0) / (75 kg + 70 kg) = 1.090 m/s .         (2b)
To find the magnitude and direction of the final velocity, we use Pythagoras' Theorem and trigonometry,

v_f = [(v_fx)² + (v_fx)²]^½ = 2.30 m/s, and

θ = tan^-1 (|v_fy/v_fx|) = 28.3°.
The final velocity of the pair is 2.30 m/s at 28.3° above horizontal.
The force on player B has to do with the impulse acting on him and his change of momentum,
F_average = I_B/Δt                     (3)
Now force and impulse are vector quantities and the i and j components must be handled separately.

F_{ave x} = (m_Bv_Bfx - m_Bv_Bix)/Δt ,                     (3a)
F_{ave y} = (m_bv_Bfy - m_Bv_Biy)/Δt .                     (3b)

Using the given values, we find

F_{ave x} = (70 kg)[(2.025 m/s) - (2.30 m/s)] / (0.30 s) = -64.17 N ,           (4a)
F_{ave y} = (70 kg)[(1.090 m/s) - (0)] / (0.30 s) = 254.33 N .         (4b)

To find the magnitude and direction of the average force, we use Pythagoras' Theorem and trigonometry,

F_ave = [(F_x)² + (F_y)²]^½ = 262 N, and

θ = tan^-1 (|F_y/F_x|) = 75.8°.
The average force on player B is 262 N at 94.2° above horizontal.
Two opposing hockey players are racing up the ice for the puck when they collide at point A as shown in the diagram below. The first hockey player has mass 90 kg and a speed of 2.7 m/s while the other has mass 82 kg and speed 3.1 m/s. The angle in the diagram is θ = 32° . After the collision, the players remain locked together (at least until the referee forces them apart). What is the magnitude and direction of the players' velocity just after they collide?

Since the collision is totally inelastic and in two dimensions, we find that we are dealing with a vector addition problem, P_T = P₁ + P₂. First we calculate the magnitude of each player's momentum using p = mv,

P₁ = m₁v₁ = (90 kg)(2.7 m/s) = 243 kg-m/s ,

P₂ = m₂v₂ = (82 kg)(3.1 m/s) = 254.2 kg-m/s.

Then we find P_T by the component method,

i j

P_1x = 0 P_1y = 243

P_2x = 254.2sin(32°)
= 134.705 P_2y = 254.2cos(32°)
= 215.574

P_Tx = 134.705 P_Ty = 458.574

Using the Pythagorean formula we find,

P_T = [(P_Tx)² + (P_Ty)²]^½ = [(134.705)² + (458.574)²]^½ = 477.949 kg-m/s .

Using trigonometry, we find the angle from

θ = arctan(P_Ty/P_Tx) = arctan(458.574/134.705) = 73.63°.

So the total momentum of the two players is P_T = (478,73.6°).
Now P_T = (m₁ + m₂)v_f, so the final velocity must be in the same direction as the total momentum. The magnitude of the velocity is

v_f = P_T/(m₁ + m₂) = 477.949 kg-m/s/ (90 kg + 82 kg) = 2.78 m/s .

So the final velocity of the two players just after the collision is v_f = (2.78 m/s, 73.6°).

Angular Momentum
Determine the direction of the angular momentum for the following cases:
Angular momentum is defined as the cross product of position and momentum, L = r × p. The direction of the angular momentum is perpendicular to the plane formed by the position and momentum vectors. For this problem that means either into the paper, denoted by ×, or out of the paper, ☉. To find the direction, we sweep our right hand through the smallest angle formed by the vector. The way the thumb points indicates the direction of the angular momentum.
Calculate the angular momentum for the following particles. Find the angle between the position and the momentum vectors.
(a) r= <4, -5, 3> and p = <1, 4, -2>
(b) r = <1, -2, 3> and p = <7, -1, 1>
(c) r = <0, 2, 0> and p = <1, 0, 0>

The angular momentum can be found either by evaluating the determinant or by using L = rpsinφ. We will use the first method to find L. We can find the angle between the momentum and position vectors using φ = sin^-1(L/rp). We find the magnitudeof the vectors using the 3D form of the Pythagorean Theorem.

(a)

L = [(-2)²+ (11)² + (21)²]^½ = 23.791 kg-m²/s

r = [(4)²+ (-5)² + (3)²]^½ = 7.0711 m

p = [(1)²+ (4)² + (-2)²]^½ = 4.5826 kg-m/s

φ = sin^-1(L/rp) = sin^-1[ 23.791 / (7.0711 × 4.5826)] = 47.2°

(b)

L = [(1)²+ (20)² + (13)²]^½ = 23.875 kg-m²/s

r = [(1)²+ (-2)² + (3)²]^½ = 3.7417 m

p = [(7)²+ (-1)² + (1)²]^½ = 7.1414 kg-m/s

φ = sin^-1(L/rp) = sin^-1[ 23.875 / (3.7417 × 7.1414)] = 63.3°

(c)

L = [(0)²+ (0)² + (-2)²]^½ = 2 kg-m²/s

r = [(0)²+ (2)² + (0)²]^½ = 2 m

p = [(1)²+ (0)² + (0)²]^½ = 1 kg-m/s

φ = sin^-1(L/rp) = sin^-1[ 2 / (2 × 1)] = 90°
Calculate the angular momentum of a phonograph record (LP) rotating at 33¹/₃ rev/min. An LP has a radius of 15 cm and a mass of 150 g. A typical phonograph can accelerate an LP from rest to its final speed in 0.35 s, what average torque would be exerted on the LP?
We are given the shape of the record. The angular momentum of a rotating body is L = Iω. An LP is a solid disk. Consulting a table of moments of inertia, we find I = ½MR². The angular velocity must be converted to rad/s

ω = 100/3 rev/min × 2p rad / rev × 1 min / 60 s = 3.4907 rad/s .
Thus we find the angular momentum of the LP to be

L = Iω = ½MR²ω = ½(0.15 kg)(0.15 m)²(3.4907 rad/s) = 5.8905 × 10^-3 kg-m²/s .
Torque is equal to the change in angular momentum with time

τ = ΔL / Δt = (L_f - L_i) / Δt = (5.8905 × 10^-3 kg-m²/s - 0) / 0.35 s = 1.68 × 10^-2 N-m .
A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless surface when two more identical non-rotating cylinders fall on top of the first. Because of friction between the cylinders they will eventually all come to rotate at the same rate. What is this final angular velocity?
We have a collision that results in a change in rotation, and the shapes are specified, so we conserve angular momentum,

L_f = L_i . (1)
The objects are rotating so their angular momentum is given by L = Iω. Thus in this particular case, equation (1) becomes

3Iω_f = Iω_i .
Solving, we find

ω_f = ω_i / 3 = 1.33 rad/s .
In a nightmare you dream that you are a hamster running in an exercise wheel. Typical hamsters are 300 g and can run at speeds of 3.2 m/s. A typical exercise wheel has a moment of inertia about its centre of 0.250 kg-m². How fast should the wheel have been rotating in your dream? The radius of the wheel is 12.0 cm. Treat the hamster as a point mass. Hint what was the angular momentum of the system before the hamster started running?

Before the hamster starts running, the exercise wheel is not rotating. Considered as a system, angular momentum must be conserved.

L_f = L_i .
For this particular problem, we know the shape of the wheel but not that of the hamster. We will treat the hamster as a particle.

L_wheel + L_hamster = 0 .
The wheel is a rotating object so its angular momentum is given by L_wheel = -Iω , where the minus sign indicates that it is into the paper. For a point particle, the angular momentum is L_hamster = Rmv out of the paper. Thus we have

-Iω + Rmv = 0 .
So the angular velocity of the wheel is

ω = Rmv / I = (0.3 kg)(0.12 m)(3.2 m/s) / (0.25 kg-m²/s) = 0.461 rad/s .
A door with width L = 1.0 m and mass M = 15 kg is hinged on one side so that it can rotate freely. A bullet, as shown, is fired into the exact centre of the door. The bullet has mass 25 g and a speed of 400 m/s. What is the angular velocity of the door with respect to the hinge just after the bullets embeds itself in the door? The door may be treated as a thin rectangular sheet. The bullet may be treated as a point mass.

Since we have a collision in which there is a change in rotation, we apply the Law of Conservation of Angular Momentum,

L_f = L_i . (1)
We are not given the shape of the bullet, so we treat it as a particle. Initially the bullet is traveling in a straight line so its angular momentum is bmv, where b is the distance of closest approach to the point of rotation. Since everything will rotate about the door hinge, we take the hinge as the point of rotation. Hence b = ½L. After the collision, the bullet is stuck in the door and rotates with the door in a circle. The angular momentum of an object moving in a circle is r²mω, where r is the radius of rotation. Clearly r = ½L.

Initially the door is not rotating and thus has no angular momentum. Afterwards, it is rotating and thus has an angular momentum given by Iω since we know it's shape. Note that the door is not rotating about its centre of mass, so we need to use the parallel axis theorem. Consulting a table, we find I_cm = (1/12)ML². The centre of mass is d = ½L from the hinge.

Thus equation (1) applied to this problem is

(½L)²mω + [(1/12)ML² + M (½L)²]ω = ½Lmv .
Dividing through by ½L and solving for ω, we find

ω = mv / [½m + (2/3)M]L = (0.025)(400) / [½(0.025) + (2/3)15] = 0.999 rad/s .
A 150-g piece of playdough slides across a frictionless table at v = 5.50 m/s. It collides with a disk of radius R = 35.0 cm and mass M = 2.50 kg which has a fixed frictionless axle. The playdough stick to the disk. Treat the playdough as a point mass. (a) If the disk is not rotating initially, what is its angular velocity after the collision? (b) What angular velocity would the disk need to have initially, if the disk stopped completely after the collision?

(a) Since we have a collision in which there is a change in rotation, we apply the Law of Conservation of Angular Momentum,

L_f = L_i . (1)
We don't know the shape of the playdough, so we will treat it as a particle. Initially the playdough is travelling in a straight line so its angular momentum is bmv, where b is the distance of closest approach to the point of rotation. Since everything will rotate about the centre of the disk, we take the centre as the point of rotation. Thus b = 0.20 m. After the collision, the playdough is stuck on the disk and rotates with the door in a circle. We know the shape of the disk and the angular momentum of a particle (the playdough) moving in a circle is r²mω, where r is the radius of rotation. Clearly r = R, the radius of the disk.

Initially the disk is not rotating and thus has no angular momentum. Afterwards, it is rotating and thus has an angular momentum given by Iω. Consulting a table of moments of inertia, we find I = ½MR².
Thus equation (1) applied to this problem is

R²mω + ½MR²ω = bmv .
Solving for ω, we find

ω = bmv / [m + ½M]R² = (0.2)(0.15)(5.5) / [(0.15) + ½(2.5)](0.35)² = 0.962 rad/s .
(b) If everything stops, L_f = 0. Equation (1) becomes

0 = bmv + ½MR²ω _initial .
Solving for ω_initial, we get

ω_initial = 2bmv / MR² = 2(0.2)(0.15)(5.5) / (2.5)(0.35)² = -1.08 rad/s .
The minus sign indicates that the disk would have to rotate clockwise.
A 22-g bug crawls from the centre to the outside edge of a 150-g disk of radius 15.0 cm. The disk was rotating at 11.0 rad/s. What will be its final angular velocity? Treat the bug as a point mass.

There will be a change in rotation as the bug moves, so we use the Law of Conservation of Angular Momentum

L_f = L_i . (1)
Assuming that the bug doesn't slip then it rotates at the same velocity as the disk. Thus bug rotates in a circle. We don't know the shape of the bug, so we treat it as a particle. The angular momentum of a particle moving in a circle is r²mω, where r is the radius of rotation. At the centre of the disk r = 0, initially. At the final position, the edge of the disk, r = R.

We know the shape of the disk. The disk is rotating and thus has an angular momentum by Iω. Consulting a table of moments of inertia, we find I = ½MR².

Thus equation (1) applied to this problem is

R²mω_f + ½MR²ω_f = (0)²mω_i + ½MR²ω_i .
Dividing through by ½R² and solving for ω_f, we find

ω_f = Mω_i / (2m + M) = (0.15)(11.0 rad/s) / [2(0.022) + 0.150] = 8.51 rad/s .
A cylindrical rod of radius r = 2.00 cm and mass 1.25 kg is upright on the edge of a rotating disk of mass 10.0 kg and radius 25.0 cm as is shown in diagram (a) below. The system is rotating at 15.0 rad/s. The rod falls on its side as shown in diagram (b). Diagrams (c) and (d) present a side view. What is the new angular velocity of the system? How much work was done in changing the shape of the object?

As the system changes shape, there will be a change in rotation, so we use the Law of Conservation of Angular Momentum

L_f = L_i . (1)
We know the shape of all the objects. Initially the cylindrical rod is a small disk rotating about the centre of the big disk The angular momentum of a rotating object is Iω. However, note that the cylindrical rod is not rotating about its own centre of mass. We must use the parallel axis theorem with d = R - r. Consulting a table of moments of inertia, the moment for a small disk is I_cm = ½mr². When the cylindrical rod falls, it is still a rotating object and its angular momentum is given by Iω, but now is has the shape of a rod. Consulting a table of moments of inertia, we find I =(1/12)mL². According to the diagram, L = 2R = 0.50 m

In both cases the disk is rotating and thus has an angular momentum by Iω. Consulting a table of moments of inertia, we find I = ½MR².

Thus equation (1) applied to this problem is

(1/12)mL²ω_f + ½MR²ω_f = [½mr² + m(R-r)²]ω₀ + ½MR²ω₀ .
Solving for ω_f, we find

ω_f = {m[½r² + (R-r)²] + ½MR²}ω₀ / {(1/12)mL² + ½MR²} .
Substituting in the given values, we get

ω_f ={[(1.25)[½(0.02)² + (0.25 − 0.02)²] + ½(10)(0.25)²}(15) / {¹/₁₂(1.25)(0.5)² + ½(10)(0.25)²} = 16.78 rad/s.
We know that that W = ΔE_system. Here the system is the two objects. The objects are rotating and the rotational kinetic energy of a rotating object is KE_rotational = ½ω² and we already know the moments of inertia of the objects and the angular velocities.
W = ½[I_rod + I_{big disk}]ω_f² − ½[I_{small disk} + I_{big disk}]ω₀²
W = ½[(1/12)mL² + ½MR²]ω_f² − ½[(½mr² + m(R-r)²) + ½MR²]ω₀²
W = ½[¹/₁₂(1.25)(0.5)² + ½(10)(0.25)²](16.78)² − ½[(1.25){½(0.02)² + (0.25 − 0.02)²} + ½(10)(0.25)²](15)² = 5.08 J

Questions? mike.coombes@kpu.ca

i		j
P_1x	= 0	P_1y	= 243
P_2x	= 254.2sin(32°) = 134.705	P_2y	= 254.2cos(32°) = 215.574
P_Tx	= 134.705	P_Ty	= 458.574