A uniform circular metal disk has radius 39 cm and mass 350 g and its center is

Question

A uniform circular metal disk has radius 39 cm and mass 350 g and its center is at the origin. Then a circular hole of radius 13 cm is cut out of it. The center of the hole is a distance 19.5 cm from the center of the disk. The disk is rotating at a rate of 2.6 rad/sec. 1) What is the moment of inertia of the modified disk about the origin? 2) What is the total linear velocity of a piece metal at the disks outer edge? 3) What is the total linear velocity of a piece metal 11.5 cm from the center? 4) What is the total linear acceleration of a piece metal at the disk s outer edge? 5) What is the total linear acceleration of a piece metal 11.5 cm from the center?

Explanation / Answer

(1) As we know, for a uniform disk

Moment of Inertia, I=.5*m*r^2

The moments of inertia are like mass, they sum. IN this case, there will be a difference sine a hole is cut out of the disk.

If the disk did not have a hole, then
I=0.5*350*39^2
= 266175 g cm^2

Again, as per the condition, hole is removed.
As the first answer post noted, you apply the paralllel axis theorm which is
Iz=Icm+m*d^2
where Iz is the moment about the axis of interest
Icm is the moment about the center of mass of the object
m is the mass
and d is the distance between the two axis

First, let's compute the mass of the hole as if it were there
the area of the large disk is
*39^2
and the area of the hole is
*13^2

the mass of the hole is
13^2*350/39^2 = 38.9 g

And, the Icm=0.5*38.9*13^2
Iz= 0.5*38.9*13^2+38.9*19.5^2
= 3287.05 + 14791.7 = 18078.8 g cm^2

So Moment of Inertia of the modified disk has
I=266175-18078.8 = 248096.2 g cm^2

Convert this value in kg-m^2 -

I = 248096.2 x 10^-7 kg-m^2 = 2.48 x 10^-2 kg-m^2.

(2) Linear velocity at the outer edge, v = w*r = 2.6*0.39 = 1.014 m/s

(3) Total linear velocity at 11.5 cm from the center -

v = 2.6*0.115 = 0.299 m/s

(4) Since the disk is rotating at a uniform speed of 2.6 rad/s. Ans it has not started from rest.

So, it has only radial acceleration.

So, total linear acceleration = its radial acceleration = a = w^2*r = 2.6^2*0.39 = 2.64 m/s^2

(5) Toal linear acceleration = 2.6^2*0.115 = 0.78 m/s^2

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