When you start your computer, the hard disk begins to spin. it takes 6 seconds of constant angular acceleration to reach full speed, at which time the computer can begin to access it. if you wanted the disk drive to reach full speed in only 2 seconds, how much more torque would the disk drive’s motor have to exert on it during the starting process?

Let:

L be the angular momentum,

Ƭ be the torque,

I be the moment of inertia , the axis of rotation is taken to be through the center of mass,

ω be the angular velocity

α is constant angular acceleration

Initially, it takes 6 s to reach full speed and we can express the torque as :

Ƭ = dL/dt => Ƭ= I α = I( Δω/Δt ) , for Δt= 6s

In order to improve access to hard disk the torque has to change according to

Ƭ”= I α” = I( Δω/Δt” ) ,for Δt”= 2s

Thus

Ƭ”/ Ƭ = ( I α”)/ (I α ) => Ƭ”/ Ƭ= Δt/ Δt” => Ƭ”/ Ƭ=6/2

Ƭ”= 3 Ƭ

The motor would have to exert a torque the magnitude of which is three times that of the original torque

f=ma newton

f=100x2.5=250n

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