Rotational Dynamics: Disk Crash! A hard drive consists of a rigid circular platt
ID: 1490244 • Letter: R
Question
Rotational Dynamics: Disk Crash!
A hard drive consists of a rigid circular platter (or disk) of radius R and thickness d, with a mass of M.
In normal operating conditions, the hard drive spins at angular velocity ?0. The read/write head is a light metallic element that floats just above the surface of the platter and both senses and imposes the magnetic structures that embody the digital data stored on the disk. The head moves radially in and out across the disk to write data in concentric circular tracks that fill the surface of the platter. When the head crashes, it scrapes against the surface of the platter, destroying the disk.
During the crash the read/write head scrapes against the disk with a coefficient of kinetic friction of ?k and normal force of N. Assume that just before the head crash the disk is rotating at ?0 radians/second. You can ignore any friction at the bearing.
For this situation, the frictional force Ff due to the read/write head is perpendicular to the radius vector, so that the torque has a magnitude ? = rFf where r is the distance of the head from the disk axis. The moment of inertia of a uniform disk of radius R and mass M is I = 1?2MR2.
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a) What is the angular acceleration associated with the torque from the crashed disk head? Put your answer in terms of r, R, ?k , N, M, ?0, and/or d as needed.
b) Assuming the crash starts at time t = 0, what is the time tstop at which the disk stops? Put your answer in terms of r, R, ?k , N, M, ?0, and/or d as needed.
c) How many revolutions does the disk undergo after the head crashes but before it stops spinning? Put your answer in terms of r, R, ?k , N, M, ?0, and/or d as needed.
d) How much energy is dissipated as heat as the head rubs against the surface of the platter while the system slows to a halt (Calculate this using both ?KE=Work and Work =?? and compare the answers)? Does this energy depend on where the head hits the disk?
Disk Plater Head Arm Head actuator Chassis http:www.computerhope.comExplanation / Answer
Radius = R
Thickness = d
Mass = M
angular velocity = 0
coefficient kinetic friction = k
normal force= N
Ff= k N
torque = R* Ff
moment of inertia of disk I= 1/2mr^2
a) alpha= Torque/I
alpha= (r * Ff)/0.5MR^2
b) = 0 + alpha(t)
0 = 0 + alpha(t)
t = 0/alpha
c) angle = -1/2*alpha*t^2 + 0*t
mind the sign: alpha is in opposite direction from 0
and since t = 0/alpha
angle = -1/2*alpha*( 0/alpha)^2 + 0* 0/alpha
angle = -1/2 0^2/apha + 0^2/alpha
angle = 1/2 0^2/alpha
rev = angle/2pi
d) K = 1/2*I* 0^2
So no it does not depend on where the heads hit the disc. It only depends on how much rotational energy the disc had when it was spinning.
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