Answer:
the correct one is C
Explanation:
To find the answer, let's propose the solution of the problem
We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.
Initial instant. Before starting to walk
L₀ = 0
Final moment. After the child is walking
L_f = I₁ w₁ + m r v₂
where index 1 is used for the platform and index 2 for the child
linear and angular velocity are related
v₂ = w₂ r
angular momentum is conserved
0 = I₁ w₁ + m r (w₂ r)
w₁ = [tex]- \frac{m r^2}{I1} \ w_2[/tex]
the moment of inertia of the platform bringing it closer to a disk or cylinder
I₁ = [tex]\frac{1}{2}[/tex] M r²
sustitute
w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]
W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]
from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds
when examining the answers the correct one is C
Answer:
Explanation:
relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system
so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero
so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong
option B is wrong because relative to ground their angular momentum should be equal and opposite
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