Question

a flywheel increases it's speed from 30 r. p. m to 60 r.p.m in 10seconds.calculate the angular acceleration and no of revolution made by the in these 10seconds​

Answer 1

Explanation:

Given a flywheel increases it's speed from 30 r. p. m to 60 r.p.m in 10 seconds.calculate the angular acceleration and no of revolution made by it in these 10 seconds

• We need to find the angular acceleration and number of revolution.
• So we need to find ω initial and ω final
• So ω i = 30 rpm (rotation per minute)
•            = 30 x 2π / 60 rad / sec
•           = π rad / sec
• So ω f = 60 rpm
•           = 60 x 2π / 60
•           = 2π rad / sec
• Also Δt = 10 secs
• Now using uniformly accelerated motion we have
• So ωf = ωi + αΔt
•     2π = π + α (10)
•  Or π = 10 α
• Or α = π / 10 rad / s^2
• We need to find the number of revolutions.
• So we have third equation of motion to find theta, so we get
•  So ωf^2 – ωi^2 = 2αΔθ
•     (2π)^2 – (π)^2 = 2 π / 10 x Δθ
•           3π^2 = π / 5 Δθ
•      Or Δθ = 15 rad
• Now number of revolutions n = Δθ / 2π
•                                                 = 15 π / 2π
•                                                 = 7.5 revolutions.

Reference link will be

https://brainly.in/question/7815311

Answer 2

Given:

A flywheel increases it's speed from 30 r. p. m to 60 r. p. m in 10sec.

To Find:

Calculate the angular acceleration and no of revolution made by the in these 10sec.

Solution:

since, we know that-

$\omega_{f}=\omega_{i}+at$

where, $\omega_{f}$ is the final angular velocity.

$\omega_{i}$ is initial angular velocity.

ω = angular acceleration , t = time

therefore, on putting given values , we get

$60=30+a\times10\\\\a=3 \ \dfrac{ rotation}{min^2}$

Now, we know that , each rotation consist of 2π radian.

therefore, no. of revolution made by flywheel in these 10 seconds is

$a = 6\pi \ \dfrac{radian}{min^2}$ .