# Rotation of a magnetic computer disk

In summary: I will definitely be a lot smarter from now on if I can just get help from the experts.In summary, the dot has traveled 1.45 revolutions since it started rotating.

Hey guys, does anyone know how to approach these two problems ... can u please help me ?

Question 1. A magnetic computer disk 8.0 am in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at 600 rad/s2 for ½ s, then coasts at a steady angular velocity for another ½ s. What is the speed of the dot at t=1.0 s ? Through how many revolutions has it turned?

Question 2. A 300 g ball and a 600 g ball are connected by a 40-cm-long massless, rigid rod. The structure rotates about its center of mass at 100rpm. What is the rotational kinetic energy ?

I'll be really really thankful if you guys tell me how to approach these problems step by step.

Second, you must show some of your own work in order to get our help. How would you start question -1- ?

As berkeman says, you'll have to do some work on your own and show where you got stuck to get help. Here are some hints:

(1) Understand http://canario.iqm.unicamp.br/MATDID/HyperPhysics/hbase/mi.html#rlin"; this is a rotational kinematics problem.
(2) What's the definition of rotational KE? What's the center of mass? What's the rotational inertia?

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I have a similar problem to this. The numbers are slightly different, though.

A 260 g ball and a 510 g ball are connected by a 37.0-cm-long massless, rigid rod. The structure rotates about its center of mass at 140 rpm.

What is its rotational kinetic energy?

Well, I know I will need to find the center of mass first.

((.260kg)(0m) + (.510kg)(.37m))/(.26kg + .51kg) = .245 m

Knowing the center of mass, then I can find I

I= (m1)(r1^2)+(m2)(r2^2)
I=(.269kg)(-.245^2)+(.510kg)((.37-.245)^2)= .024 kgm^2

Now, I am not sure where I need to go from here.

What's the definition of rotational kinetic energy?

KE= (1/2)mv^2

Samurai Weck said:
KE= (1/2)mv^2
That's translational KE. What's the corresponding formula for rotational KE?

Samurai Weck said:
KE= (1/2)mv^2

That's the kinetic energy for linear motion. What is it for rotational motion? Hint -- just as mass comes into play for linear KE, the "moment of inertia" comes into play for rotational KE...

Ooooooh. K=(1/2)Iw^2

Krot= (1/2)(.024kgm^2)(14.66 rad/s)^2 = 2.53 J

Thanks A LOT. This was the one problem out of my homework that was kicking my behind.

## 1. What is the purpose of rotation in a magnetic computer disk?

The rotation of a magnetic computer disk is necessary to read and write data on the disk. The disk contains a magnetic coating that stores data as magnetic patterns, and the rotation allows for the disk to pass under the read/write head, which reads and writes the data.

## 2. How does the rotation of a magnetic computer disk work?

The rotation of a magnetic computer disk is typically powered by a motor that spins the disk at a constant speed. The motor is controlled by a drive circuit, which also controls the read/write head. As the disk rotates, the read/write head moves back and forth across the surface of the disk to access different areas and read or write data.

## 3. What factors affect the rotation speed of a magnetic computer disk?

The rotation speed of a magnetic computer disk is affected by the motor speed, the size and weight of the disk, and any external forces such as friction or air resistance. The rotation speed is typically measured in revolutions per minute (RPM) and can range from a few thousand to tens of thousands of RPM.

## 4. Can the rotation of a magnetic computer disk be adjusted?

Yes, the rotation speed of a magnetic computer disk can be adjusted by the drive circuit and motor. This allows for different data transfer rates and can also help with reducing noise or vibrations within the disk.

## 5. How does the rotation of a magnetic computer disk affect data storage and retrieval?

The rotation of a magnetic computer disk is essential for data storage and retrieval. Without rotation, the read/write head would not be able to access different areas of the disk and data could not be read or written. The speed and precision of the rotation also play a role in the speed and accuracy of data storage and retrieval.

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