A high-speed drill rotating ccw at 2400rpmcomes to a halt in 2.5s.

a. What is the magnitude of the drill’s angular acceleration?

b. How many revolutions does it make as it stops?

Short Answer

Expert verified

a. The magnitude of the drill’s angular acceleration-100rad/s2.

b. 50revit make as it stops.

Step by step solution

01

Given information

ωi=2400rpm

ωf=0

t=2.5s

02

Part a) Step 1: Calculation

To convert angular velocity from revolution per minutes to the revolution per second:

ωi=2400revmin×1min60s×2πrad1revωi=251rad/s

To calculate angular acceleration:

α=ωf-ωitα=0rad/s-251rad/s2.5sα=-100rad/s2
03

Part b) Step 1: Calculation

To calculate number of revolution

2θ=(ωf+ωi)tθ=(ωf+ωi)t2θ=(0rad/s+251rad/s)(2.5s)2θ=(251rad/s)(2.5s)2θ=313.75rad

To convert radian in revolutions

θ=2πNN=θ2πN=313.75rad(2)(3.14rad)N=49.96revN=50rev

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