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)
@ anonymous aha, you instead were casually around here...
so... no ideas for this problem???
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)
and the rotation?
To determine the amount of energy lost in this collision, we can use the principle of conservation of linear momentum. The momentum before the collision is equal to the momentum after the collision.
First, let's calculate the initial angular velocity of the merry-go-round using the principle of conservation of angular momentum. The angular momentum is given by the equation:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of a uniform disk is given by the equation:
I = (1/2) * M * R^2
where M is the mass of the disk and R is the radius.
Plugging in the values, we have:
I = (1/2) * 200 kg * (110 cm)^2
= (1/2) * 200 kg * (1.1 m)^2
= 12100 kg*m^2
The initial angular momentum is then:
L_initial = I * ω_initial
Next, we need to calculate the final angular momentum after the collision. Since the hammer is not moving after hitting the merry-go-round, the system of the merry-go-round and the hammer has zero angular momentum.
Therefore, we can equate the initial and final angular momentum:
L_initial = L_final
I * ω_initial = I * ω_final
Simplifying, we find:
ω_final = ω_initial
Now, let's calculate the kinetic energy before and after the collision. The kinetic energy of a rotating body is given by the equation:
KE = (1/2) * I * ω^2
The initial kinetic energy is:
KE_initial = (1/2) * I * ω_initial^2
The final kinetic energy is:
KE_final = (1/2) * I * ω_final^2
Since ω_final = ω_initial (as determined earlier), the final kinetic energy is the same as the initial kinetic energy:
KE_final = KE_initial
Therefore, there is no change in kinetic energy during this collision. As a result, the loss of energy (|ΔE|) is equal to zero Joules.
Hence, the amount of energy lost in this collision is 0 Joules.