A person of mass 73 kg stands at the center of a rotating merry-go-round platfor
ID: 1789146 • Letter: A
Question
A person of mass 73 kg stands at the center of a rotating merry-go-round platform of radius 2.6 m and moment of inertia 810 kgm2 . The platform rotates without friction with angular velocity 0.95 rad/s . The person walks radially to the edge of the platform.
Part A
Calculate the angular velocity when the person reaches the edge.
Express your answer using three significant figures and include the appropriate units.
For Part A, I got 0.590 rad/s. Which was correct.
Part B
Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.
Express your answers using two significant figures. Enter your answers numerically separated by a comma.
For part B I got 365.5 as my inital and 274.9 as my final. Are these wrong? I put them in as two significant figures each but it is stilling being counted wrong. Help with this would be much appreciated.
Explanation / Answer
(a) Apply conservation of angular momentum:
I1 * w1 = I2 * w2
Its moment of inertia is:
I = I-disk + m*r²
Where I-disk is the given moment of inertia for the platform, m is the man's mass and r is the man's distance from the platform's axis of rotation. Thus:
I1 = I-disk + (73 kg)(0m)² = 810 kg-m²
I2 = I-disk + (73 kg)(2.6m)² = 1303.5 kg-m²
w1 = 0.95 rad/s
Now from the above expression of angular momentum, the final angular velocity is:
w2 = (I1 / I2) * w1
w2 = (810 / 1303.5) * (0.95 rad/s)
w2 = 0.59 rad/s
(b) The kinetic energy of a rotating body is simply:
KE = (1/2) * I * w²
Since we know the moment of inertia of the system for each case, along with its corresponding angular velocity, we can find the kinetic energy for both situations as:
KE1 = (1/2)(810 kg-m²)(0.95 rad/s)²
KE1 = 365.6 J
KE2 = (1/2)(1303.5 kg-m²)(0.59 rad/s)²
KE2 = 226.9 J
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