Question

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

Answer 1

19.49 rev

Step-by-step 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

Answer 2

`t=10.87s`

Step-by-step explanation:

Angular acceleration constant - Angular kinematics apply

`w=w_(i)+at\\`

The initial angular speed is

`w_(i)=215(rev)/(min)((2\pi rad)/(1rev) )((1min)/(60.0s) ) \\w_(i)=22.5rad/s`

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

`t=(w_(f)-w_(i))/(\alpha ) \\t=(0-22.5rad/s)/(-2.07rad/s^2)\\ t=10.87s`