A circular drum of radius 40 cm is initially rotating at 400 rev/min. It is brought to a stop after making 50 revolutions. What is the angular acceleration and the stopping time? (2.79 rad/s², 150 s)


Given:

Radius of the circular drum = r = 40 cm = 0.4 m
Initial angular velocity = Ѡi = 400 rev/min = `\frac {400 ｘ 2 π rad}{60 s}` = `\frac {40}{3}`π rad s⁻¹
Final angular velocity = Ѡf = 0 rad s⁻¹
angular displacement = 𝚹 = 50 rev = 50 ｘ 2 π rad = 100 π rad

To Find:

Angular acceleration = 𝛂 = ?
Stopping time = t = ?

Solution:

Using the third equation of motion for angular motion

2 𝛂 𝚹 = Ѡf ² -  Ѡi²

putting values

2 ｘ 𝛂 ｘ 1oo π rad = (0 rad s⁻¹)² -  (`\frac {40}{3}` π rad s⁻¹)²

200 𝛂  π rad = (0 rad s⁻¹) ² -  `\frac {1600}{9}` π² rad²s⁻²

by simplifying we get

𝛂  = - 2.79 rads⁻²-------------Ans. 1

-ve shows the deceleration


Now we will find time t by using the first equation of motion for angular motion, 

Ѡf  =  Ѡi + 𝛂 t

or

t = `\frac {Ѡf  - Ѡi}{𝛂}`

by putting values

t = `\frac {o rad s⁻¹ - 40/3 π rad s⁻¹}{- 2.79 rads⁻²}`

t = `\frac {- 40/3 π rad s⁻¹}{- 2.79 rad s⁻²}`

t = `\frac {40 π rad s⁻¹}{ 3 ｘ 2.79 rad s⁻²}`

t = 15 s ------------------Ans. 2


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