A particle of mass m=0.1kg and speed v0=5m/s collides and sticks to theend of s

Question

A particle of mass m=0.1kg and speed v0=5m/s collides and sticks to theend of s solid cylinder of mass M=1.0 kg and radius R = 20cm. If the cylinder is initially at rest and is pivoted about a frictionless axle through its center, what is the final angular velocity (in rads/s) of the system after the collision? Answer is 4.2 but how do I get there? A particle of mass m=0.1kg and speed v0=5m/s collides and sticks to theend of s solid cylinder of mass M=1.0 kg and radius R = 20cm. If the cylinder is initially at rest and is pivoted about a frictionless axle through its center, what is the final angular velocity (in rads/s) of the system after the collision? Answer is 4.2 but how do I get there? Answer is 4.2 but how do I get there?

Explanation / Answer

Now look at the angular momentum after the collision. This will be the sum of the momentums of the cylinder and the particle.

Lafter = Lcyl + Lpart

Now each of these depends on the angular speed and the corresponding moment of inertia.
Lcyl = Icyl*w and Lpart = Ipart*w ..... w = angular speed
Lafter = Icyl*w + Ipart*w
Lafter = w*(Icyl + Ipart)
w = Lafter / (Icyl + Ipart)
w = Lbefore / (Icyl + Ipart) ..... since Lafter = Lbefore

For a solid cylinder with a central axis the moment of inertia is:
Icyl = (1/2)*m*r^2 where m = 1.0 kg and r = 20 cm = 0.2 m
Icyl = (1/2)*1.0*(0.2)^2
Icyl = 0.02 kg m^2

For the particle it is just:
Ipart = m*r^2 where m = 0.1 kg and r = 0.2 m
Ipart = 0.1*(0.2)^2
Ipart = 0.004 kg m^2

Icyl + Ipart = 0.02 + 0.004 = 0.024
w = Lbefore / (Icyl + Ipart) = 0.1/0.024
w = 4.2 rad/s

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