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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32

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.

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