A small circular block of mass M traveling with a speed v on a frictionless table collides and sticks to the end of a thin rod of with length D and mass M. The picture shows a top down view of the block and rod on the frictionless table. What is the rod's angular velocity after the collision? Express your answer in some combination of M, v, and D.
m*v_cf+m*ω*d/2?
im not sure tbh
Answer
To find the rod's angular velocity after the collision, we can use the principle of conservation of angular momentum.
Angular momentum (L) is given by the product of moment of inertia (I) and angular velocity (ω): L = Iω.
Before the collision, only the block has angular momentum. Since it is moving in a circular path about its center, the moment of inertia of the block (I_block) can be expressed as: I_block = ½MR^2, where R is the radius of the block.
After the collision, the block and rod move together as a single system, which we will call the block-rod system.
The moment of inertia of the entire block-rod system (I_block-rod) can be expressed as: I_block-rod = I_block + I_rod, where I_rod is the moment of inertia of the rod.
The moment of inertia of the rod about its axis is given by: I_rod = ⅓ML^2, where L is the length of the rod.
Since the block sticks to the end of the rod, the total mass (M) of the block-rod system is 2M (M for the block and M for the rod).
Using the principle of conservation of angular momentum, the angular momentum before the collision (L_before) is equal to the angular momentum after the collision (L_after).
L_before = L_block = I_block × ω_before,
L_after = L_block-rod = I_block-rod × ω_after.
Since the block is initially moving in a circular path, its initial angular velocity (ω_before) can be calculated as v/R, where v is its linear velocity. Conversely, the angular velocity of the block-rod system after the collision (ω_after) can be expressed as R × ω_after.
Using the conservation of angular momentum, we can set up the equation as:
I_block × ω_before = I_block-rod × ω_after.
Substituting the expressions for I_block and I_block-rod:
(½MR^2) × (v/R) = (½MR^2 + ⅓ML^2) × (R × ω_after).
Simplifying the equation:
(v/R) = [(2/3) + (R^2/L^2)] × (R × ω_after).
Now, we can solve for ω_after:
ω_after = (v/R) / [(2/3) + (R^2/L^2)].
So, the rod's angular velocity after the collision can be expressed as (v/R) / [(2/3) + (R^2/L^2)], where v is the speed of the block, R is the radius of the block, and L is the length of the rod.