A horizontal 790-N merry-go-round of radius 1.30 m is started from rest by a constant horizontal force of 50 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder.)
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To find the kinetic energy of the merry-go-round, we need to determine its angular acceleration and rotational kinetic energy. Here's how we can do it step by step:
Step 1: Calculate the angular acceleration
The angular acceleration can be found using the formula:
α = τ / I
Where α is the angular acceleration, τ is the applied torque, and I is the moment of inertia.
We are given that the applied torque is equal to the force applied tangentially multiplied by the radius:
τ = F * r
Plugging in the values, we get:
τ = 50 N * 1.30 m = 65 N·m
The moment of inertia for a solid cylinder rotating about its central axis is given by:
I = (1/2) * m * r^2
Where m is the mass of the merry-go-round. We can calculate the mass using the given force and the acceleration due to gravity (9.8 m/s^2):
m = F / g
m = 50 N / 9.8 m/s^2 ≈ 5.10 kg
Now, substituting the values into the formula for the moment of inertia:
I = (1/2) * 5.10 kg * (1.30 m)^2 = 4.17 kg·m^2
Finally, substituting the values of τ and I into the formula for angular acceleration:
α = 65 N·m / 4.17 kg·m^2 ≈ 15.60 rad/s^2
Step 2: Calculate the final angular velocity
The final angular velocity can be determined using the following equation:
ω = ω₀ + α * t
Where ω is the final angular velocity, ω₀ is the initial angular velocity (which is zero since the merry-go-round starts from rest), α is the angular acceleration, and t is the time.
Substituting the values:
ω = 0 + (15.60 rad/s^2) * 2.0 s ≈ 31.20 rad/s
Step 3: Calculate the rotational kinetic energy
The rotational kinetic energy of the merry-go-round can be calculated using the formula:
KE = (1/2) * I * ω^2
Substituting the values:
KE = (1/2) * 4.17 kg·m^2 * (31.20 rad/s)^2 ≈ 2051 J
Therefore, the kinetic energy of the merry-go-round after 2.0 s is approximately 2051 Joules.