I'm reposting this one because I'm still stuck...
A uniform rod of mass M and length d is initially at rest on a horizontal and frictionless table contained in the xy plane, the plane of the screen. The rod is free to rotate about an axis perpendicular to the plane and passing through the pivot point at a distance d/3 measured from one of its ends. A small point mass m, moving with speed v0, hits the rod and stick to it at the point of impact at a distance d/3 from the pivot.
a. If the mass of the rod is M=4m, what is the magnitude of the angular velocity of the rod+small mass system after the collision?
b. Using again M=4m. What is the speed of the center of mass of the rod right after collision?
I know I need conservation of angular momentum, but how do I set up the equation -- i.e what is the initial moment of inertia expression, and what is the final?
angular momentum of system before collision about pivot point
= m v R = m Vo (d/3) = m Vo d/3
Moment of inertia of rod about pivot = (1/12)(4m) d^2 +(4m)(d/6)^2
= 4 m d^2 (1/12 + 1/36) =4 m d^2(4/36) = (4/9)m d^2
Moment of inertia of m about pivot = m (d/3)^2
Total moment of inertia after crash
= (4/9 + 1/9)m d^2 = (5/9) m d^2
if at angular velocity w
total angular momentum = I w = (5/9) m d^2 w
so
(5/9) m d^2 w = m Vo d/3
solve for w
for part b, v = w times distance from pivot which is d/6
Why did the point mass go to therapy? Because it had a lot of angular momentum issues!
Let's break this problem down step by step. First, we need to find the initial and final moment of inertia expressions.
The initial moment of inertia (Ii) for the system is the sum of the moment of inertia of the rod (Ir) and the moment of inertia of the small mass (Im).
For a uniform rod rotating about one end perpendicular to its length, the moment of inertia is given by:
Ir = (1/3) * M * (d/3)^2
Since the small mass is glued to the rod at a distance d/3 from the pivot, its moment of inertia is:
Im = m * (d/3)^2
Now, let's move on to the final moment of inertia (If). After the point mass sticks to the rod, the system becomes a compound body with a new moment of inertia.
Since the mass is sticking to the rod, we can treat them as one rigid body. The moment of inertia of this compound body can be found using the parallel axis theorem.
The parallel axis theorem states that If = Ir + Im + M * (d/3)^2
Now that we have the initial and final moment of inertia expressions, let's apply the conservation of angular momentum.
Angular momentum is defined as L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since there are no external torques acting on the system, the total angular momentum before and after the collision must be conserved.
Before the collision, the initial angular momentum (Li) is given by:
Li = Ii * ωi
After the collision, the final angular momentum is:
Lf = If * ωf
Since angular momentum is conserved, Li = Lf. So we can set up the equation:
Ii * ωi = If * ωf
Now, let's plug in the expressions for Ii and If:
(1/3) * M * (d/3)^2 * ωi = [(1/3) * M * (d/3)^2 + m * (d/3)^2 + M * (d/3)^2] * ωf
Simplifying the equation, we get:
(1/3) * M * (d/3)^2 * ωi = [(1 + 1/3 + 1) * M + m] * (d/3)^2 * ωf
Now, let's solve for ωf, which is the angular velocity of the system after the collision:
ωf = [(1/3) * M * (d/3)^2 * ωi] / [(1 + 1/3 + 1) * M + m]
For part a, since we know the values of M and m, we can substitute them and calculate the magnitude of ωf using the given values.
Finally, for part b, we need to calculate the speed of the center of mass after the collision. The speed of the center of mass (v_cm) is given by v_cm = R * ω, where R is the distance from the axis of rotation to the center of mass.
In this case, R = (2/3) * d, since the center of mass of the rod is located at (2/3) of its length.
Using the value of ωf calculated in part a, we can plug it into the equation to find the speed of the center of mass.
To solve this problem, we can use the law of conservation of angular momentum. The initial angular momentum before the collision is equal to the final angular momentum after the collision.
a. To find the magnitude of the angular velocity of the rod+small mass system after the collision, we need to calculate the initial and final angular momenta.
Before the collision:
The initial moment of inertia of the rod about the pivot point is given by I1 = (1/3)Md^2 (for a rod rotating about one end).
The initial angular momentum of the rod is L1 = I1 * ω1, where ω1 is the initial angular velocity of the rod.
After the collision:
The final moment of inertia of the system (rod + small mass) about the pivot point is given by I2 = (1/3)Md^2 + md^2 (since the small mass sticks to the rod).
The final angular momentum of the system is L2 = I2 * ω2, where ω2 is the final angular velocity.
By applying the conservation of angular momentum, we have L1 = L2, which gives:
I1 * ω1 = I2 * ω2
Substituting the expressions for I1 and I2, we have:
(1/3)Md^2 * ω1 = [(1/3)Md^2 + md^2] * ω2
Simplifying the equation, we get:
(1/3)Mω1 = [(1/3)M + m]ω2
Since the small mass m sticks to the rod, their angular velocities will be the same after the collision. Thus, we can write:
ω1 = ω2
Substituting this back into the equation, we have:
(1/3)Mω1 = [(1/3)M + m]ω1
Canceling ω1 from both sides of the equation, we get:
(1/3)M = (1/3)M + m
Simplifying further, we find:
m = 0
This implies that the equation cannot be satisfied with the given values. Therefore, there might be an error in the problem statement, or some other information is missing.
b. Since we couldn't solve part a, we cannot move on to part b since it requires knowing the final angular velocity of the system.
To solve this problem, we can use the principle of conservation of angular momentum. Remember that angular momentum is the product of angular velocity and moment of inertia.
a. To find the magnitude of the angular velocity of the rod+small mass system after the collision, we first need to determine the initial and final moments of inertia.
The initial moment of inertia (Ii) of the rod can be calculated using the formula for a rod rotating about its center:
Ii = (1/12) * M * d^2
When the small mass m collides with the rod, it sticks to it and becomes a part of the system. The final moment of inertia (If) can be calculated by considering the rod and the small mass as a combined object, which can be treated as a point mass located at the center of mass of the system. The distance of the small mass from the pivot point is d/3, so the final moment of inertia (If) is:
If = Ii + m * (d/3)^2
The principle of conservation of angular momentum states that the initial angular momentum (Li) equals the final angular momentum (Lf). Before the collision, the rod is at rest, so its initial angular momentum is zero. After the collision, the system will rotate with an angular velocity ω, so the final angular momentum (Lf) is given by:
Lf = If * ω
Setting Li = Lf, we have:
0 = If * ω
From this equation, we can solve for the magnitude of the angular velocity (ω) of the rod+small mass system after the collision.
b. To find the speed of the center of mass of the rod right after the collision, we can use the principle of conservation of linear momentum. The linear momentum of the system is conserved in the absence of any external forces.
The initial linear momentum (Pi) is given by the mass of the small mass m times its initial velocity v0. The final linear momentum (Pf) is given by the combined mass of the rod and the small mass (M+m) times the velocity of the center of mass of the system.
Since the rod is initially at rest, Pi = 0. After the collision, the speed of the center of mass (Vf) is the same as the speed of the small mass m (v0). Thus, we have:
Pf = (M+m) * Vf
Using the conservation of linear momentum, we set Pi = Pf and solve for Vf. This will give us the speed of the center of mass of the rod right after the collision.
Now you have the equations necessary to solve for the magnitude of the angular velocity and the speed of the center of mass of the rod in parts a and b, respectively.