# Rotation of a Rigid Body

1. Nov 2, 2006

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.

2. Nov 2, 2006

### Staff: Mentor

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

3. Nov 2, 2006

### Staff: Mentor

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?

Last edited by a moderator: Apr 22, 2017
4. Dec 7, 2009

### 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.

5. Dec 7, 2009

### Staff: Mentor

What's the definition of rotational kinetic energy?

6. Dec 7, 2009

### Samurai Weck

KE= (1/2)mv^2

7. Dec 7, 2009

### Staff: Mentor

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

8. Dec 7, 2009

### Staff: Mentor

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...

9. Dec 7, 2009

### Samurai Weck

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