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A 1.99 kg toy train rolls around a circular horizontal track. If the train has an angular acceleration of -1.16 rads/s2 and is released with an angular speed of 19.5 rpm, what time is required for the train to come to a complete stop? 1.76 s 16.8 s 161 s 0.568 s

Question

A 1.99 kg toy train rolls around a circular horizontal track. If the train has an angular acceleration of -1.16 rads/s2 and is released with an angular speed of 19.5 rpm, what time is required for the train to come to a complete stop? 1.76 s 16.8 s 161 s 0.568 s

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Solution

To solve this problem, we need to use the formula for angular acceleration, which is:

Δω = α * Δt

Where: Δω is the change in angular speed, α is the angular acceleration, and Δt is the change in time.

We know that the train is released with an angular speed of 19.5 rpm, but we need this in rad/s. To convert rpm to rad/s, we multiply by 2π/60. So, the initial angular speed (ωi) is:

ωi = 19.5 rpm * 2π/60 ≈ 2.04 rad/s

The train comes to a complete stop, so the final angular speed (ωf) is 0 rad/s. Therefore, the change in angular speed is:

Δω = ωf - ωi = 0 rad/s - 2.04 rad/s = -2.04 rad/s

The angular acceleration is given as -1.16 rad/s². We can now solve for Δt:

Δt = Δω / α = -2.04 rad/s / -1.16 rad/s² ≈ 1.76 s

So, the time required for the train to come to a complete stop is approximately 1.76 seconds.

This problem has been solved

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