Respuesta :
Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Explanation:
Given that;
height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum = 2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ = [tex]\frac{1}{2}[/tex] mv²
v² = 2gh₁
v = √(2gh₁)
from the second equation; s = ut + [tex]\frac{1}{2}[/tex] at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )
we plug in our values
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{5}[/tex]mR²) ω²
v = √( [tex]\frac{10}{7}[/tex]gh₁ )
so we substitute √( [tex]\frac{10}{7}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( [tex]\frac{10}{7}[/tex]gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c)
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{3}[/tex]mR²) ω²
v = √( [tex]\frac{6}{5}[/tex]gh₁ )
so we substitute √( [tex]\frac{6}{5}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √( [tex]\frac{6}{5}[/tex]gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{5}[/tex]mR²) ω²
v = √( [tex]\frac{10}{7}[/tex]gh₁ )
so we substitute √( [tex]\frac{10}{7}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( [tex]\frac{10}{7}[/tex]gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
We have that for the Question it can be said that
- [tex]A solid steel sphere sliding down the ramp without friction = 1.55m[/tex]
- [tex]A solid steel sphere rolling down the ramp without slipping = 1.309m[/tex]
- A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping =[tex]1.2m[/tex]
- A solid aluminum sphere rolling down the ramp without slipping = [tex]1.309m[/tex]
From the question we are told
Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R= 15mm in each case; assume that the density of steel is 7.8 g/cm3; and assume that the density of aluminum is 2.7 g/cm3.
Generally the equation for sliding without friction is mathematically given as
[tex]V = \sqrt{4Hh}[/tex]
the equation for sliding without slipping is mathematically given as
[tex]X = \sqrt{\frac{4Hh}{1+I/mR^2}}[/tex]
A) A solid steel sphere sliding down the ramp without friction.[tex]V = \sqrt{4*1.5*0.4}\\\\= 1.55m[/tex]
B) .A solid steel sphere rollingdown the ramp without slipping.[tex]I = 2/5 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m[/tex]
C) A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping.
[tex]I = 2/3 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/3}}\\\\= 1.2m[/tex]
D) A solid aluminum sphere rolling down the ramp without slipping.
[tex]X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m[/tex]
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