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KonstantinChe [14]
3 years ago
11

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.
Physics
2 answers:
Gekata [30.6K]3 years ago
7 0

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

Inga [223]3 years ago
3 0

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

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