Two 2.3 kg balls are attached to the ends of a thin rod of negligible mass, 65 cm in length. The rod is free to rotate in a vertical plane about a horizontal axis through its center. With the rod initially horizontal as shown, a 44 gm wad of wet putty drops onto one of the balls with a speed of 3.7 m/sec and sticks to it.
A) What is the ratio of the kinetic energy of the entire system just after the collision to just before the collision?
B)Through what angle will the system rotate until it momentarily stops?
a)
m=putty
M=balls
Li=Lf
mvr=Iw
w=mvr/I
Ki=1/2*m*v^2
Kf=Iw^2=1/2*m^2*v^2*r^2/I
Things that cancel:
1/2
m
v^2
Left over:
m*r^2/I << THIS is the answer. Plug in I
I=M*(L/2)+(M+m)*(L/2)=L/2*(2M+m)
To solve this problem, we can apply the principle of conservation of angular momentum.
Let's start with part A of the question, where we need to find the ratio of the kinetic energy of the entire system just after the collision to just before the collision.
1) Compute the initial angular momentum (L) just before the collision. The initial angular momentum is given by the equation:
L_initial = m1 * v1 * r1 + m2 * v2 * r2
Where:
m1 and m2 are the masses of the balls (2.3 kg in this case)
v1 is the initial speed of the wad of wet putty (3.7 m/s in this case)
r1 is the distance of the impact point from the axis of rotation (half the length of the rod, so 0.5 * 65 cm = 0.325 m)
v2 is the tangential speed of the balls due to the rotation of the rod (initially 0 since they are not rotating yet)
r2 is the radius of rotation of the balls (the length of the rod divided by 2, so 65 cm/2 = 0.325 m)
Calculating the initial angular momentum:
L_initial = (0.044 kg)(3.7 m/s)(0.325 m) + (2.3 kg)(0 m/s)(0.325 m)
L_initial = 0.05699375 kg*m^2/s
2) After the collision, the entire system (balls + wad of putty) will start rotating. Since the wet putty sticks to one of the balls, the total mass of the system is now the sum of the masses of both balls and the wad of putty (2 balls each weighing 2.3 kg plus the putty weighing 0.044 kg).
Total mass after collision = 2.3 kg + 2.3 kg + 0.044 kg = 4.644 kg
3) Compute the final angular momentum (L_final) just after the collision. The final angular momentum can be calculated using the equation:
L_final = I * ω
Where:
I is the moment of inertia of the entire system
ω is the angular velocity of the system
Calculating the moment of inertia of the system:
I = m1 * r1^2 + m2 * r2^2 + m3 * r3^2
Where:
m1 and m2 are the masses of the balls (2.3 kg each)
m3 is the mass of the wad of putty (0.044 kg)
r1 and r2 are the distances of the balls from the axis of rotation (r2 = 0 since the putty sticks to one ball, and r1 is the same as before, 0.325 m)
r3 is the distance of the putty from the axis of rotation (the length of the rod, so 0.65 m)
Calculating the moment of inertia:
I = (2.3 kg)(0.325 m)^2 + (2.3 kg)(0)^2 + (0.044 kg)(0.65 m)^2
I = 0.30001203 kg*m^2
4) The system rotates about a horizontal axis through its center, so we can write the angular momentum as:
L = I * ω
Since the putty sticks to one of the balls, they will rotate together. They have the same angular velocity, so we can write:
L_final = I * ω_final
5) We know that the angular momentum is conserved, so:
L_initial = L_final
And we can find the final angular velocity (ω_final) by rearranging the equation:
L_final = L_initial
I * ω_final = L_initial
ω_final = L_initial / I
ω_final = 0.05699375 kg*m^2/s / 0.30001203 kg*m^2
ω_final = 0.18996 rad/s
6) Now, we can calculate the final kinetic energy (K_final) of the system just after the collision. The final kinetic energy is given by the equation:
K_final = 0.5 * I * ω_final^2
Calculating the final kinetic energy:
K_final = 0.5 * 0.30001203 kg*m^2 * (0.18996 rad/s)^2
K_final = 0.005412599 J
7) Finally, we calculate the initial kinetic energy (K_initial) of the system just before the collision. The initial kinetic energy can be calculated using the equation:
K_initial = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2
Where:
m1 and m2 are the masses of the balls (2.3 kg each)
v1 is the initial speed of the wad of wet putty (3.7 m/s in this case)
v2 is the tangential speed of the balls due to the rotation of the rod (initially 0 since they are not rotating yet)
Calculating the initial kinetic energy:
K_initial = 0.5 * (2.3 kg)(3.7 m/s)^2 + 0.5 * (2.3 kg)(0 m/s)^2
K_initial = 19.3045 J
8) Now, we can find the ratio of the kinetic energy of the entire system just after the collision to just before the collision:
Ratio = K_final / K_initial
Ratio = 0.005412599 J / 19.3045 J
Ratio ≈ 0.000280719
Therefore, the ratio of the kinetic energy of the entire system just after the collision to just before the collision is approximately 0.000280719.
Now, let's move on to part B of the question, which asks for the angle through which the system rotates until it momentarily stops.
We can use the principle of conservation of mechanical energy to solve this.
1) At the moment the system momentarily stops, all of its initial kinetic energy will have been converted into potential energy. Therefore, we can equate the initial kinetic energy (K_initial) to the gravitational potential energy (PE) at the topmost point.
K_initial = PE
2) The gravitational potential energy at the topmost point is given by the equation:
PE = m * g * h
Where:
m is the total mass of the system (4.644 kg)
g is the acceleration due to gravity (9.8 m/s^2)
h is the height of the topmost point, which is the length of the rod (65 cm or 0.65 m)
Calculating the gravitational potential energy:
PE = (4.644 kg)(9.8 m/s^2)(0.65 m)
PE = 29.41346 J
3) Equating the initial kinetic energy to the gravitational potential energy:
K_initial = PE
19.3045 J = 29.41346 J
4) Finally, we can solve for the angle through which the system rotates until it momentarily stops. The angle (θ) can be calculated using the equation:
θ = arcsin(K_initial / PE)
Calculating the angle:
θ = arcsin(19.3045 J / 29.41346 J)
θ ≈ 0.708 rad
Therefore, the system will rotate through an angle of approximately 0.708 radians until it momentarily stops.