Kinetic Energy of Rotating Solid disk or cylinder about symmetry axis

[Schulze]

Homework Statement

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

Homework Equations

I of MGR = 1/2mR²
angular KE = 1/2Iw²
F = ma
linear KE = 1/2mv²
τ=αI = FR

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?

 

[maajdl]

Of course, linear KE = 0 .

The axis of the MGR is fixed.
If the axis would not be fixed, you "angular+linear" calculation would be wrong anyway.
You would then need to specify the problem more precisely: what would be the meaning of a tangential force?
How could a tangential force accelerate the center of mass and put the MGR into rotation both together at the rate would indicated?