An electric motor rotating a workshop grinding wheel at a rate of 215 rev/min is switched off. Assume constant angular deceleration of magnitude 2.07 rad/s 2 . Through how many revolutions does the wheel turn before it finally comes to rest

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Khoso123 Khoso123 · Answer 1 · 2020-03-19T04:01:15+01:00

Answer:

[tex]t=10.87s[/tex]

Explanation:

Angular acceleration constant - Angular kinematics apply

[tex]w=w_{i}+at\\[/tex]

The initial angular speed is

[tex]w_{i}=215\frac{rev}{min}(\frac{2\pi rad}{1rev} )(\frac{1min}{60.0s} ) \\w_{i}=22.5rad/s[/tex]

The time to stop (reach a final speed of wf=0) with α= -2.07 rad/s² is:

[tex]t=\frac{w_{f}-w_{i}}{\alpha } \\t=\frac{0-22.5rad/s}{-2.07rad/s^2}\\ t=10.87s[/tex]

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asuwafohjames asuwafohjames · Answer 2 · 2020-03-19T08:14:50+01:00

Answer:

19.49 rev

Explanation:

Using,

ω₂² = ω₁²+2αθ.................................. Equation 1

Where ω₂ = Final angular velocity, ω₁ = Initial angular velocity, α = angular deceleration, θ = number of revolution.

make θ the subject of the equation

θ = (ω₂²- ω₁²)/2α............................. Equation 2

Given: ω₂ = 0 rad/s (comes to rest), ω₁ = 215 rev/min = rad/s, α = -2.07 rad/s²(deceleration) = -2.07(572.957) = -1186.02 rev/min²

Substitute into equation 2

θ = (0²-215²)/2(-1186.02)

θ = -46225/-2372.04

θ = 19.49 rev

Hence the wheel makes 19.49 rev