As seen from above, a playground carousel is rotating counterclockwise about its

Question

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of 1.60 m, an initial angular speed of 3.18 rad/s, and a moment of inertia of 128 kg · m2. The mass of the person is 43.8 kg. Find the final angular speed of the carousel after the person climbs aboard. f =

Explanation / Answer

I assume we are to ignore any torques the person applied to the carousel to get on board...this would change the angular momentum of the carousel..

so making this assumption, we will use conservation of angular momentum
angular momentum = I w where I is the moment of inertia and w is the angular velocity
call I0, w0 the values of moment of inertia and ang vel before the person gets on board and I, w the values after cons of angular momentum tells us that

I0 w0 = I w ==========> w=(I0 w0))/I

we are given I0 and w0, so we need to find the moment of inertia after the person climbs on board think of the person as a point particle of mass M and a distance R fromthe rotation axis...this particle contributes an amount of moment of inertia equal to MR^2...so the person adds 43.8kg x (1.6m)^2 = 112.13 kgm^2 to the value of I

so we have:
I0=128 kgm^2
w0=3.18 rad/s
I=128kgm^2+112.13kgm^2 = 240.13 kg m^2

and w = (128)(3.18)/240.13=1.695 rad/s

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