A high-speed drill reaches 2000rpmin0.50s .

a. What is the drill's angular acceleration?

b. Through how many revolutions does it turn during this first 0.50s?

Short Answer

Expert verified

(a) The angular acceleration is419rad/s2

(b) The number of revolution is 8.30rev.

Step by step solution

01

Step 1. Given information

(a) The change in angular velocity,Δω=2000revmin

(b) The change in time interval, Δt=0.50s

The expression for angular displacement in a given time interval is,

θ(t)=ωiΔt+12α(Δt)2

Here, ωiis the initial velocity, Δtis the time interval and is angular acceleration.

02

Part (a)

The angular acceleration.

α=ΔωΔtα=2000rpm2π1rotation1min60s0.5s=418.6rad/s2α419rad/s2

03

Part (b)

The number of revolutions turned during the first is given by,

θ(t)=ωiΔt+12αΔt2θ(t)=(0)Δt+12419rad/s20.5s2θ(t)=52.4rad1rev2πrad=8.30rev

Therefore, the number of revolution0.5s in is8.30rev .

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