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2 years ago

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A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. What is the angular acceleration? B. How man

y radians did the disk turn while stopping ? C. how many revolutions?

1 answer:

Ne4ueva [31]2 years ago

6 0

Answer:

A. α = - 1.047 rad/s²

B. θ = 14.1 rad

C. θ = 2.24 rev

Explanation:

A.

We can use the first equation of motion to find the acceleration:

$\omega_f = \omega_i + \alpha t$

where,

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s

t = time = 3 s

α = angular acceleration = ?

Therefore,

$0\ rad/s = 3.14\ rad/s + \alpha(3\ s)$

<u>α = - 1.047 rad/s²</u>

B.

We can use the second equation of motion to find the angular distance:

$\theta = \omega_it +\frac{1}{2}\alpha t^2\\\theta = (3.14\ rad/s)(3\ s) + \frac{1}{2}(1.04\ rad/s^2)(3)^2$

<u>θ = 14.1 rad</u>

C.

θ = (14.1 rad)(1 rev/2π rad)

<u>θ = 2.24 rev</u>

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