Answer:
4 smaller disks
Explanation:
We are given;
Mass of smaller and larger disks = M
Radius of smaller disk = R
Radius of larger disk = 4R
Formula for moment of inertia about cylinder axis is:
I = ½MR²
Thus;
For small disk, I_small = ½MR²
For large disk, I_large = ½M(2R)² = 2MR²
We are told that moment of inertia of System A consists of two of the larger disks. Thus;
I_A = 2 × I_large = 2 × 2MR²
I_A = 4MR²
We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;
I_B = I_large + n(I_small)
Where n is the number of smaller disks.
I_B = 2MR² + n(½MR²)
I_B = MR²(2 + n/2)
We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;
I_A = I_B
So;
4MR² = MR²(2 + n/2)
MR² will cancel out to give;
4 = 2 + n/2
Multiply through by 2 to give;
8 = 4 + n
n = 8 - 4
n = 4
