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Answer :
The kinetic energy of the merry-go-round after 3.93 seconds is approximately 135.7 J.
Here's the approach:
Calculate the moment of inertia (I) of the merry-go-round.
We can use the formula for the moment of inertia of a solid cylinder:
I = (1/2) × M × [tex]R^2[/tex]
where:
I is the moment of inertia (kg m²)
M is the mass of the merry-go-round (kg)
R is the radius of the merry-go-round (m)
First, we need to convert the weight of the merry-go-round from Newtons (N) to mass (kg) using the following conversion:
M = weight / g
where:
M is the mass (kg)
weight is the weight of the merry-go-round (N)
g is the acceleration due to gravity (9.8 m/s²)
Therefore:
M = 652 N / 9.8 m/s² ≈ 66.43 kg
Now, we can calculate the moment of inertia:
I = (1/2) × 66.43 kg × (1.37 m)² ≈ 58.49 kg m²
Calculate the angular acceleration (α) of the merry-go-round.
We can use Newton's second law of rotational motion, which states:
τ = I × α
where:
τ is the torque (N m)
I is the moment of inertia (kg m²)
α is the angular acceleration (rad/s²)
In this case, the torque is equal to the applied tangential force (F_t) multiplied by the radius (R):
τ = F_t × R
Therefore:
α = (F_t × R) / I
α = (77 N × 1.37 m) / 58.49 kg m² ≈ 1.91 rad/s²
Calculate the angular velocity (ω) of the merry-go-round after 3.93 seconds.
We can use the following kinematic equation for rotational motion:
ω = α × t
where:
ω is the angular velocity (rad/s)
α is the angular acceleration (rad/s²)
t is the time (s)
Therefore:
ω = 1.91 rad/s² × 3.93 s ≈ 7.51 rad/s
Calculate the kinetic energy (KE) of the merry-go-round.
We can use the formula for the kinetic energy of a rotating object:
KE = (1/2) × I × ω^2
where:
KE is the kinetic energy (J)
I is the moment of inertia (kg m²)
ω is the angular velocity (rad/s)
Therefore:
KE = (1/2) × 58.49 kg m² × (7.51 rad/s)² ≈ 135.7 J
Therefore, the kinetic energy of the merry-go-round after 3.93 seconds is approximately 135.7 J.
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