A merry-go-round in a playground has a mass of 200 kg and radius of 1.50 m. You apply a force F = 23 N tangentially to the edge of the merry-go-round for a time of 6.5 s. The moment of inertia of the merry-go-round is 1/2(mR2).
What is the angular acceleration of the merry-go-round?
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Use Kr=.5 I w f^2
To find the angular acceleration of the merry-go-round, we can use the equation:
τ = I * α
where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.
First, let's calculate the torque applied. The torque is given by:
τ = F * r
where F is the force applied tangentially to the edge of the merry-go-round, and r is the radius of the merry-go-round.
τ = 23 N * 1.50 m
τ = 34.5 Nm
Next, we need to determine the moment of inertia of the merry-go-round. According to the given information, the moment of inertia is given by:
I = 1/2(m * R^2)
where m is the mass of the merry-go-round and R is its radius.
I = 1/2 * (200 kg) * (1.50 m)^2
I = 225 kgm^2
Now, we can rearrange the torque equation to solve for the angular acceleration:
α = τ / I
α = 34.5 Nm / 225 kgm^2
α ≈ 0.153 rad/s^2
Therefore, the angular acceleration of the merry-go-round is approximately 0.153 rad/s^2.