Answer:
(a) Angular acceleration will be [tex]100.53rad/sec^2[/tex]
(b) 50 revolution
Explanation:
We have given time t = 2.5 sec
Initial speed of the drill [tex]\omega _0=0rad/sec[/tex]
Speed after 2.5 sec [tex]=2400rpm=\frac{2\pi \times 2400}{60}=251.327rad/sec[/tex]
From first equation of motion we know that
[tex]\omega =\omega _0+\alpha t[/tex]
[tex]251.327 =0+\alpha \times 2.5[/tex]
[tex]\alpha =100.53rad/sec^2[/tex]
(b) From second equation of motion we know that
[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2[/tex]
[tex]\Theta =0\times 2.5+\frac{1}{2}\times 100.53\times 2.5^2=314.16rad=\frac{314.16}{2\pi }=50revolution[/tex]
