Arunner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. the runner's velocity relative to the earth has magnitude 3.10m/s . the turntable is rotating in the opposite direction with an angular velocity of magnitude 0.210rad/s relative to the earth. the radius of the turntable is 3.10m , and its moment of inertia about the axis of rotation is 81.0kg? m2

question: find the final angular velocity of the system if the runner comes to rest relative to the turntable. (you can treat the runner as a particle.)

-0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m  

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of  the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec  

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec  

Solving for ωf, we get:

ωf = -0.84 rad/sec  (clockwise)

the answer is a)

mark brainliest

not too sure thinking all accelerations are pointed down