Coin Rolling Without Slipping: Physics Final Study Question

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To solve the problem of a coin rolling without slipping, the relevant formula is ωf² = ωi² + 2αΔθ, where ωf is the final angular speed, ωi is the initial angular speed, α is the angular acceleration, and Δθ is the angular displacement. Given the coin's initial angular speed of 18.4 rad/s and an angular acceleration of -2.17 rad/s², the formula can be rearranged to find Δθ. After calculating Δθ, the distance the coin rolls can be determined by multiplying Δθ by the radius of the coin, which is 1.15 cm. This approach effectively combines angular motion principles to find the total distance traveled before the coin comes to rest.
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A coin with a diameter of 2.30 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 18.4 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 2.17 rad/s2, how far does the coin roll before coming to rest?


Can anyone help- I'm trying to study for a physics final and I'm getting nowhere! Thanks.
 
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Unless I'm missing something here, you can use this formula:

vf2 = vi2 + 2aΔx

Dividing both sides by R2, the coin's radius squared, we get:

ωf2 = ωi2 + 2αΔθ

Where α is the coin's angular acceleration, which is negative in this case. From here you can find Δθ and multiplying by R you can find the total distance traveled by the coin.
 
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Great...thanks :biggrin:
 
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