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Chapter 12: Q. 1 (page 330)

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,螖蝇=2000revmin

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

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

胃(t)=蝇i螖t+12伪(螖t)2

Here, 蝇iis the initial velocity, 螖tis the time interval and is angular acceleration.

02

Part (a)

The angular acceleration.

伪=螖蝇螖t伪=2000rpm2蟺1rotation1min60s0.5s=418.6rad/s2伪鈮�419rad/s2

03

Part (b)

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

胃(t)=蝇i螖t+12伪螖t2胃(t)=(0)螖t+12419rad/s20.5s2胃(t)=52.4rad1rev2蟺rad=8.30rev

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

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