Question

Before the invention of a wheel turning on an axle, ancient people moved heavy loads by placing rollers under them. (Modern people use rollers, too. Any hardware store will sell you a roller bearing for a lazy susan.) A stone block of mass 920 kg moves forward at 0.415 m/s, supported by two uniform cylindrical tree trunks, each of mass 82.0 kg and radius 0.343 m. No slipping occurs between the block and the rollers or between the rollers and the ground. Find the total kinetic energy of the moving objects.

Answer 1

Answer:

The total kinetic energy of the moving objects is 84.52 J

Explanation:

The total energy of the moving objects is given as;

Total energy = Kinetic Energy of the stone block + Kinetic Energy of the rollers + Rotational Kinetic Energy of the rollers

∑E = 0.5·m₁·v₁² + 2×0.5·m₂·v₂² + 2×0.5·I·ω²

Where:

m₁ = Mass of the stone block = 920 kg

v₁ = Velocity of the stone block = 0.415 m/s

m₂ = Mass of tree trunk rollers = 82.0 kg each

v₂ = Velocity of the rollers = v₁/2 = (0.415 m/s)/2 = 0.2075 m/s

I = Moment of inertia of the rollers, (solid cylinder bout its axes) = m₂·r²/2

ω = Angular velocity of the rollers = v₂/r

r = Radius of the tree trunk rollers = 0.343 m

Therefore

∑E = 0.5×920 kg×(0.415 m/s)² + 2×0.5×82.0 kg×(0.2075 m/s)² + 2×0.5·m₂·r²/2·ω²

82.754 + 2×0.5·82.0 kg·(0.343 m)²/2·((0.2075 m/s)/0.343 m)²

= 82.754 +1.765 = 84.52 J

∑E = Total Energy = 84.52 J

Therefore the total Kinetic Energy = 84.52 J.