Question

An electric drill starts from rest and undergoes uniform angular acceleration for a period of 0.145 s until it turns at a rate of 2.85 ✕ 10^4 revs/min.
(a) What is the drill's angular acceleration (in rad/s^2)?
(b) What is the angle (in radians) through which the drill rotates during this period?

Answer 1

Answer

given,

time period, t = 0.145 s

angular velocity, ω = 2.85 x 10⁴ rev/ min

               $\omega =2.85\times \dfrac{2 \pi}{60}$

               $\omega =2251.47\ rad/s$

using rotational motion equation

$\omega =\omega _{o}+\alpha t$

Since it starts from rest ,initial angular velocity ωo=0

final angular velocity

$\alpha=\dfrac{\omega}{t}$

$\alpha=\dfrac{2251.47}{0.145}$

α = 15,527 rad/s²

b)

again using equation of rotational motion

$\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}$

$\theta =\frac{1}{2}\times 15527\times 0.145^{2}$

    θ = 163.23 rad