1. The problem statement, all variables and given/known data A horizontal 845.0 N merry-go-round with a radius of 4.3 m is started from rest by a constant horizontal force of 74.0 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.2 s. Assume it is a solid cylinder. The acceleration of gravity is 9.81 m/s2 2. Relevant equations I of MGR = 1/2mR² angular KE = 1/2Iw2 F = ma linear KE = 1/2mv² τ=αI = FR 3. The attempt at a solution weight of MGR = 845 N mass of MGR = 845/9.81 = 86.1366 kg R = 4.3 m I of MGR = 1/2mR² = (0.5)(86.1366 kg)(4.3 m)² = 796.3328 kgm² applied tangential force (F) = 74 N time F is applied = t = 2.2 s Torque applied = (74 N)(4.3 m) = Iα α = 318.2 N·m/796.3328 kgm² = 0.39958 rad/s² w = αt = (0.39958 rad/s²)(2.2 s) = 0.8790797 rad/s angular KE = 1/2Iw² = (0.5)(796.3328 kgm²)(0.8790797 rad/s)² = 307.695468 J F = ma a = F/m = 74 N/86.1366 kg = 0.85910054 m/s² v = at = (.85910054 m/s² )(2.2 s) = 1.8900212 m/s linear KE = 1/2mv² = (0.5)(86.1366 kg)(1.8900212 m/s)² = 153.8477 J TOTAL KE = 1/2Iw² + 1/2mv² = 307.695468 J + 153.8477 J = 461.5432 J Possible reason for missing problem: Not sure if this is correct, but I think that there is no linear KE because the center of mass of the object is not moving. Therefore, I think the correct answer is 307.695468 J. I don't know if this line of reasoning is correct or not?