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A merry-go-round (pictured) is sitting in a playground. It is free to rotate, but is currently stationary. You can model it as...Asked by annonymous
A merry-go-round (pictured) is sitting in a playground. It is free to rotate, but is currently stationary. You can model it as a uniform disk of mass 200 kg and radius 110 cm (consider the metal poles to have a negligible mass compared to the merry-go-round). The poles near the edge are 97 cm from the center.
Someone hits one of the poles with a 9 kg sledgehammer moving at 19 m/s in a direction tangent to the edge of the merry-go-round. The hammer is not moving after it hits the merry-go-round.
How much energy |ΔE| is lost in this collision? (enter a positive number for the absolute value in Joules)
Someone hits one of the poles with a 9 kg sledgehammer moving at 19 m/s in a direction tangent to the edge of the merry-go-round. The hammer is not moving after it hits the merry-go-round.
How much energy |ΔE| is lost in this collision? (enter a positive number for the absolute value in Joules)
Answers
Answered by
Anonymous
You forgot to mention that this problem is part if your 8.01 MIT exam and asking for help during your exam is a violation of the Honor Code.
Answered by
an0nym801
@ anonymous aha, you instead were casually around here...
Answered by
elli
so... no ideas for this problem???
Answered by
Greco
m_1: The mass of the merry-go-round
m_2: The mass of the sledgehammer
v_1: The velocity of the merry-go-round before the collision
v_2: The velocity of the sledgehammer before the collision
v_1': The velocity of the merry-go-round after the collision
v_2': The velocity of the sledgehammer after the collision
v_1='(m_2*v_2)/m_1 (1)
you put your v_1 in eq (2) and ready
((0.5*m_1*v_1^2)+(0.5*m_2*v_2^2))-((0.5*m_1*v_1'^2)+(0.5*m_2*v_2'^2)) (2)
m_2: The mass of the sledgehammer
v_1: The velocity of the merry-go-round before the collision
v_2: The velocity of the sledgehammer before the collision
v_1': The velocity of the merry-go-round after the collision
v_2': The velocity of the sledgehammer after the collision
v_1='(m_2*v_2)/m_1 (1)
you put your v_1 in eq (2) and ready
((0.5*m_1*v_1^2)+(0.5*m_2*v_2^2))-((0.5*m_1*v_1'^2)+(0.5*m_2*v_2'^2)) (2)
Answered by
juanpro
and the rotation?
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