Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO′ is x/4 Ma². Then, the value of x will be ____________.

Solution

Physics sir · 4 steps

The axis lies in the plane of the discs, so start with each disc's moment about its own diameter.

1. Read the figure. Each disc has radius R = a/2. The axis OO' lies in the plane and passes through the centres of the top and bottom discs. The left and right discs touch the axis, so their centres are a distance R = a/2 from it.
2. Top and bottom discs. For these two the axis is a diameter. About a diameter, a disc has 1/4MR² = Ma²/16. Together the pair gives Ma²/8.
3. Side discs. Each side disc's own diameter is parallel to OO' at distance a/2, so apply the parallel axis theorem: Ma²/16 + M(a/2)² = 5Ma²/16. Together the pair gives 5Ma²/8.

   I = Ma²/8 + 5Ma²/8 = 6Ma²/8 = 3/4Ma²

4. So. Comparing with x/4Ma² gives x = 3.

The common slip: 4. Using 1/2MR² (the axis perpendicular to the disc) instead of 1/4MR² gives a total of Ma², so x = 4.

The idea to keep. For an axis lying in the plane of a disc, use 1/4MR², not 1/2MR². Then shift it with the parallel axis theorem. Always convert a given diameter to radius first.

Answer from the NTA final answer key. Solution by PrepWiser.