Rotation of a magnetic computer disk

[ada15]

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.

 

[berkeman]

Welcome to the PF, ada15. First of all, do not double-post your questions. I deleted your duplicate thread in the Advanced Physics forum.

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

 

[Doc Al]

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?

 

[Samurai Weck]

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.

 

[Doc Al]

What's the definition of rotational kinetic energy?

 

[Samurai Weck]

KE= (1/2)mv^2

 

[Doc Al]

KE= (1/2)mv^2

That's translational KE. What's the corresponding formula for rotational KE?

 

[berkeman]

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...

 

[Samurai Weck]

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

w(omega) = 140rpm(1min/60sec)(2pi/rotation)=14.66 rad/s

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.