# Moment of inertia and a gyroscope

1. Jan 15, 2008

### Distr0

1. The problem statement, all variables and given/known data
A gyroscope consists of a uniform disc of mass M and diameter d attached at its centre to a light spindle of length l perpendicular to the plane of the disc.

i) Show that the moment of inertia, I, of the disc around the axis of the spindle is Md²/8.

2. Relevant equations

I=MR² Moment of inertia

3. The attempt at a solution

I cant understand why it becomes I=Md²/8 as R=d/2 and R²=d²/4 not d²/8

or am i missing something?

2. Jan 15, 2008

### malawi_glenn

I = MR^2/2 for a circular plate.

3. Jan 15, 2008

### Distr0

ok i'll ask my lecturer about that tomorrow as in her solutions to this she has written

I=MR²

thanks for the help

4. Jan 15, 2008

### malawi_glenn

5. Jan 15, 2008

### DylanB

The MOI of a disc rotating around its centre is certainly I = MR^2/2

the quatity MR^2 is used when applying the parallel axis theory.

6. Jan 15, 2008

### malawi_glenn

It is very hard to see how this gyroscope is built and how it rotates :)

7. Jan 16, 2008

### theinfojunkie

yea, I=MR^2 is for used on a particle or a person in some examples.
I=(MR^2)/2 is for a disc

8. Jan 17, 2008

### Shooting Star

Must be a careless mistake. Let the OP clarify with the teacher.

What sort of persons in which examples?

9. Jan 17, 2008

### Distr0

I was given two very similar questions (which I still cant tell apart)

Question 1

A uniform horizontal disc of mass M and radius R rotates about its vertical axis with angular frequency w. Find and expression for the moment of inertia.

Solution

Mass per unit area (lambda) = M/piR^2

Mass of small ring of thickness dr (dm) = lambda*2pi*r*dr

I = integral (0->R) r^2 dm
I = integral (0->R) r^2*lambda*2pi*r dr
I = 2pi*lambda integral (0->R) r^3 dr
I = 2pi*lambda*(R^4)/4
I = 2pi*(M/pi*R^2)*(R^4)/4
I = (MR^2)/2

Question 2

A gyroscope consists of a wheel of mass M and radius R attatched to a light central rod of length l that is perpendicular to the plan of the wheel. If the wheel is a uniform ring with light spokes determine an expression for the moment of inertia of the wheel about the axis of the spindle.

Solution

I = MR^2

10. Jan 17, 2008

### Shooting Star

Both correct.

What do you mean you can't tell them apart? The first is a disk, where the mass is distributed from the centre upto the circumference. The second is a ring, where all the mass is at a distance of R from the centre.