Respuesta :
Answer:
A solid moment of inertia is [tex]I = \frac{mr^2}{2}[/tex].
Here, both the solid cylinder and the cylindrical shell have the same mass, the same radius, and turn on a horizontal, friction-less axle.
The solid cylinder has less inertia than the cylindrical shell, and it requires less torque to rotate, meaning that the solid cylinder weight block falls faster than the cylindrical shell itself.
Fill in the blanks, in order.
Gravitational potential energy, Translation kinetic energy, Kinetic energy;
Hallow, Solid, Hallow, Solid;
Hallow, Transitional kinetic energy, Hallow
Answer:
The answer to the queations are;
A) The block attached to the solid cylinder would hit the ground first.
B) By the time the blocks reach the ground, they have transformed identical amounts of _gravitational potential_______energy into_____rotational kinetic________ energy of the cylinders. energy of the blocks and _______translational kinetic_____But the moment of inertia of a ____hollow______cylinder is higher than that of a ____solid_______ cylinder of the same mass, so more of the energy of the system is in the form of rotational kinetic energy for the ______hollow_____________cylinder than for the ___solid_______ one. This leaves less energy in the form of translational kinetic energy for the ____hollow________cylinder. But it is the ____translational kinetic________ energy that determines the speed of the block. So the block moves more slowly for the system with the ______hollow______cylinder, and so its block reaches the ground last.
Explanation:
To solve the question, we note that
The total energy of motion of the moving cylinders is equal to
K[tex]_{TOT[/tex] = 1/2·m·v² + 1/2·I·ω²
Where
m = Mass
v = Velocity
ω = Angular velocity
I = moment of inertia where I for hollow cylinder = MR² and
I for solid cylinder = 1/2·MR².
Therefore we have
K[tex]_{TOT[/tex] for solid cylinder = 1/2·m·v² + 1/2·I·ω² = 1/2·m·v² + 1/2·1/2·MR²·ω²
= 1/2·m·v² + 1/4·MR²·v²/r² = 1/2·m·v² + 1/4·M·v² = 3/4·m·v²
For the hollow cylinder, we have
K[tex]_{TOT[/tex] = 1/2·m·v² + 1/2·MR²·ω² = 1/2·m·v² + 1/2·MR²·v²/r² = 1/2·m·v² + 1/2·m·v²
= m·v²
From conservation of energy the initial potential energy is transformed into potential energy as follows
PE = m·g·h
Where:
m = Mas
g = Gravitational acceleration
h = height
Therefore
For the solid cylinder 3/4·m·v² = m·g·h and v² = [tex]\frac{3}{4} \frac{m*g*h}{m}[/tex] and v = [tex]\sqrt{\frac{4}{3} gh}[/tex]
For the hollow cylinder m·v² = m·g·h and v² = [tex]\frac{m*g*h}{m}[/tex] and v = [tex]\sqrt{gh}[/tex]
This shows that the solid cylinder has a higher downward velocity and the block attached to the solid cylinder would hit the ground first
B) By the time the blocks reach the ground, they have transformed identical amounts of _gravitational potential_______energy into_____rotational kinetic________ energy of the cylinders. energy of the blocks and _______translational kinetic_____But the moment of inertia of a ____hollow______cylinder is higher than that of a ____solid_______ cylinder of the same mass, so more of the energy of the system is in the form of rotational kinetic energy for the ______hollow_____________cylinder than for the ___solid_______ one. This leaves less energy in the form of translational kinetic energy for the ____hollow________cylinder. But it is the ____translational kinetic________ energy that determines the speed of the block. So the block moves more slowly for the system with the ______hollow______cylinder, and so its block reaches the ground last.