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DanielleElmas [232]
3 years ago
10

Initially a wheel rotating about a fixed axis at a constant angular deceleration of 0.5 rad/s 2 has an angular velocity of 0 rad

/s and an angular position of 6.1 rad. What is the angular position of the wheel after 3.9 s
Mathematics
1 answer:
Zigmanuir [339]3 years ago
7 0

Answer:

\theta = 5.83\ rad

Step-by-step explanation:

given,

angular deceleration, α = -0.5 rad/s²

final angular velocity,ω_f = 0 rad/s

angular position, θ = 6.1 rad

angular position at 3.9 s = ?

now, Calculating the initial angular speed

\omega_f^2 = \omega_i^2 + 2 \alpha \theta

0 = \omega_i^2 - 2\times 0.5\times 6.1

\omega_i = \sqrt{6.1}

\omega_i = 2.47\ rad/s

now, angular position calculation at t=3.9 s

\theta = \omega_i t + \dfrac{1}{2}\alpha t^2

\theta =2.47\times 3.9 - \dfrac{1}{2}\times 0.5\times 3.9^2

\theta = 5.83\ rad

Hence, the angular position of the wheel after 3.9 s is equal to 5.83 rad.

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Step-by-step explanation:

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Alternate means the angles are on different sides of the transversal (horizontal line in this case). Interior means they are between the parallel lines (lines a and b in this case). (∠2 and ∠6) and (∠3 and ∠7) are the alternate interior angles but only the first is an option.

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I honestly don't see how you can solve this without making assumptions so I will make assumptions.

Assuming the whole figure is symmetrical vertically ∠1 is congruent to the same spot on the right side of the shape. This angle would be supplementary to the angle of the angle outside of the shape adjacent on the right making the angle (which is corresponding, therefore congruent to ∠2) 133°.

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