A uniform rod AB of length 2a and mass M is freely pivoted at A and is held with B vertically above A. It is then allowed to fall and when B is vertically below A it strikes a stationary particle , also of mass M,which sticks to the rod at B. The rod then turns through a further angle alpha before coming to rest. Find alpha.
letting the zero PE system at A, then
initial PE=mga
final PE =-mga cosAlpha-MG*2a*CosAlpha
solve for cosAlpha, setting final=initialPE
CosAlpha= 1/(-1-2)=-1/3
where alpha is measured from the vertical downward, clockwise
To find the angle α through which the rod turns after the collision, we need to apply the principle of conservation of angular momentum.
The initial angular momentum of the system before collision is zero, as the rod is stationary and the particle is initially vertically above point A. After the collision, the combined system of the rod and the particle will have an angular momentum.
The angular momentum is given by the equation:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Before the collision, the rod is rotating about point A with angular velocity ω. The moment of inertia of the rod about point A is given by:
I_rod = (1/3)M(2a)^2
After the collision, the mass of the system becomes 2M (rod + particle combined). The moment of inertia of the system about point A is given by:
I_system = (1/3)(2M)(2a)^2 + M(2a)^2
Since the system is initially at rest, the total angular momentum before collision is zero. After the collision, the angular momentum can be calculated as:
L_system = I_system * ω_system
where ω_system is the angular velocity after the collision.
Since the particle sticks to the rod, the rod and the particle will rotate together as a single unit. The length of the rod is 2a, and the distance from point A to the center of mass of the system (rod + particle) is a. The angular displacement α is the angle through which the system rotates before coming to rest.
Using the principle of conservation of angular momentum:
L_rod + L_particle = L_system
0 + 0 = I_system * ω_system
(1/3)M(2a)^2 * ω_rod + M(2a)^2 * ω_particle = (1/3)(2M)(2a)^2 * ω_system
Simplifying the equation:
(1/3) * 4a^2 * ω_rod + 4a^2 * ω_particle = (2/3) * 4a^2 * ω_system
Dividing both sides by 4a^2:
(1/3) * ω_rod + ω_particle = (2/3) * ω_system
Since the particle is initially stationary, its angular velocity is zero:
(1/3) * ω_rod + 0 = (2/3) * ω_system
Multiplying both sides by 3:
ω_rod = 2 * ω_system
Substituting ω_rod into the equation:
2 * ω_system + 0 = (2/3) * ω_system
Simplifying the equation:
(2/3) * ω_system = 2 * ω_system
Multiplying both sides by (3/2):
ω_system = 4 * ω_system
Since ω_system cannot equal zero, we can divide both sides by ω_system:
1 = 4
This equation is not true, which means there is no consistent solution. Therefore, the system cannot come to rest at an angle α.