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The net reaction of the disc on the block is

m ω 2 R sin ω t j ˆ – mg k ˆ
2 1 m ω 2 R (e ω t – e – ω t ) j ˆ + mg k ˆ
2 1 m ω 2 R (e 2 ω t – e –2 ω t ) j ˆ + mg k ˆ
–m ω 2 R cos ω t j ˆ – mg k ˆ

Correct answer: 2 1 m ω 2 R (e ω t – e – ω t ) j ˆ + mg k ˆ

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