# Kinetic Energy of Rotating Solid disk or cylinder about symmetry axis

**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/s

^{2}

**2. Relevant equations**

I of MGR = 1/2mR²

angular KE = 1/2Iw^{2}

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?

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