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 = 400ｘ2πrad60s = 403π 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 -  (403 π rad s⁻¹)²

200 𝛂  π rad = (0 rad s⁻¹) ² -  16009 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 = orads⁻¹-403πrads⁻¹-2.79rads⁻²

t = -403πrads⁻¹-2.79rads⁻²

t = 40πrads⁻¹3ｘ2.79rads⁻²

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


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