Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 2R. System A consists of two of the larger disks rigidly connected to each other with a common axis of rotation. System B consists of one of the larger disks and a number of the smaller disks rigidly connected with a common axis of rotation. If the moment of inertia for system A equals the moment of inertia for system B, how many of the smaller disks are in system B? 1

number of smaller disks in system B = 4

Explanation:

Moment of Inertia of smaller disk is:

I_small = (1/2)MR²

Moment of inertia of larger disk is;

I_large = (1/2)M(2R)² = 2MR²

Now total moment of Inertia of system A will be 2 of the larger disk I_large

Thus, I_A = 2 x (2MR²) = 4MR²

While moment of inertia of system B will consist of 1 number of larger disk and n number of smaller disk.

Thus;

I_B = 2MR² + n((1/2)MR²)

Thus, I_B = MR²(2 + (n/2))

Since the two systems have equal moment of inertia, we will ewuate I_A to I_B. Thus,

I_A = I_B gives;

4MR² = MR²(2 + (n/2))

MR² will cancel out to give ;

4 = 2 + (n/2)

n/2 = 4 - 2

n/2 = 2

n = 4

protons in

prolly a beacuse the momentuem the car has increases while going down hill