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Dynamics_Part25

Dynamics_Part25 - Now for this geometry v = R and = R Thus...

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Now, for this geometry, v = ω R and ρ = R . Thus, we must solve the following equations. N cos θ μ s N sin θ = mg N sin θ + μ s N cos θ = m ω 2 R Multiplying the first equation by cos θ and the second by sin θ , the sum of the resulting equations is N D cos 2 θ + sin 2 θ i = mg cos θ + m ω 2 R sin θ N = mg cos θ + m ω 2 R sin θ Using this value of N in the vertical-force-balance equation, we find D mg cos θ + m ω 2 R sin θ i (cos θ μ s sin θ ) = mg or, dividing through by m and regrouping terms, g cos θ (cos θ μ s sin θ ) + ω 2 R sin θ (cos θ μ s sin θ ) = g Consequently, ω 2 R sin θ (cos θ μ s sin θ ) = g [1 cos θ (cos θ μ s sin θ )] = g (1 cos 2 θ + μ s sin θ cos θ ) = g (sin 2 θ + μ s sin θ cos θ ) = g sin θ (sin θ + μ s cos θ ) Therefore, the maximum rotation rate is ω max = ± g R sin θ + μ s cos θ cos
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