Respuesta :
The total kinetic energy of the moving objects in the question is;
Total Kinetic energy = 37.321 J
We are given;
Mass of stone; m_st = 672 kg
Speed of stone; v_st = 0.319 m/s
Mass of cylindrical tree trunks; m_cyl = 82 kg
radius; r = 0.343 m
Angular velocity; ω = v_wheel/r
We are told that a stone is supported by two uniform cylindrical tree trunks. Thus; v_wheel = v_st/2 = 0.319/2 = 0.1595 m/s
ω = 0.1595/0.343
ω = 0.465 m/s
The translational energy of the stone and the cylindrical system is is;
E_trans = (¹/₂m_st*(v_st)²) + 2(¹/₂m_cyl*(v_wheel)²)
E_trans = (¹/₂ × 672 × 0.319²) + (¹/₂ × 82 × 0.1595²)
E_trans = 36.278 J
Now, the rotational energy of the cylindrical system is;
E_rot = 2(¹/₂Iω²)
Where I is moment of inertia and for a cylinder it is I = ¹/₂mr²
Thus; I = ¹/₂ × 82 × 0.343²
I = 4.825 kg.m²
Then; E_rot = 2(¹/₂ * 4.825 * 0.465²)
E_rot = 1.043 J
Total Kinetic energy = E_trans + E_rot
Total Kinetic energy = 36.278 + 1.043
Total Kinetic energy = 37.321 J
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We have that for the Question, it can be said that the total kinetic energy of the moving objects is
- K.E=37.13J
From the question we are told
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 672 kg moves forward at 0.319 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.
Generally the equation for the Kinetic total is mathematically given as
[tex]\\\\K.E_T=K.E_{t}+K.E_{rot}\\\\Therefore\\\\K.E_T=0.5m_s v^2+2(0.5m_t(v/2)^2)+2(0.5I_t w^2)\\\\K.E_T=0.5(m_s+3/4m_t)v^2[/tex]
Therefore
[tex]K.E=0.5(672.8+3/4(82))(0.318)^2[/tex]
K.E=37.13J
Hence
the total kinetic energy of the moving objects is
K.E=37.13J
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