Question

A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

Answer 1

Answer:

$ΔTita = 4205.6 rad$

Explanation:

$w_{i}$ means initial angular velocity, which is 0 rev/min

$w_{f}$ means final angular velocity, which is $2.513*10^{4}rev/min$

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

$w_{f}$ = (2π/60) *2.513*10^{4} rad/s

$w_{f}$= 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

$a = 821.40 rad/s^{2}$

so the rotational motion about a fixed axis is:

$w^{2} _{f} =w^{2} _{i}$ + 2αΔTita    where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = $\frac{w^{2} _{f}-w^{2} _{i} }{2a}$

$ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40$

$ΔTita = 4205.6 rad$

Answer 2

Answer:

(a) α = 822.5rad/s²

(b) θ = 4211rad

Explanation:

Given

ω = 2.513×10⁴ rev/min = 2.513×2π/60 rad/s = 2632rad/s

t = 3.20s

ωo = 0rad/s initially at rest.

α = (ω – ωo)/t = (2632 – 0)/3.20 = 822.5rad/s²

ω² = ωo² + 2α(θ – θo)

But θo = 0rad.

2632² = 0² + 2×822.5(θ – 0)

6927424 = 1645θ

θ = 6927424/1645 = 4211.2rad