Question

A disc is rotating at a rate 1200 rpm. it is stopped bu applying a constant torque in 5 second.number pf revolution made by the disc during stopping process?​

Answer 1

Explanation:

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Answer 2

Answer:

The number of revolutions made by the disc during the stopping process = 50.

Solution:

Initial velocity, v = 1200 rpm = 1200/60 rps

                           = 20 rps (rotations per second)

Initial angular velocity, ω₁ = 2πv

                                = 2π(20)

                                = 40π rad/s

The disc is stopped by applying a torque for 5 seconds.

∴ Final angular velocity, ω₂ = 0 rad/s

Time, t = 5 seconds

Using the first equation of rotational motion, we have

ω₂ = ω₁ + αt

where α = Angular acceleration

(0) = 40π + α(5)

0 = 40π + 5α

5α = -40π

α = -40π/5

α = -8π rad/s²

Now using third equation of rotational motion, we have

ω₂² - ω₁² = 2αθ

where θ = angular displacement

(0)² - (40π)² = 2(-8π)θ

-1600π² = -16πθ

θ = (-1600π²)/(-16π)

θ = 100π rad

Number of revolutions made by the disc during stopping process, N = θ/2π

N = 100π/2π

N = 50 revolutions

Hence, 50 revolutions are made by the disc during the stopping process.

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