Question

2.(Ramp section) 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.a.A solid steel sphere slidingdown the ramp without friction.b.A solid steel sphere rollingdown the ramp without slipping.c.A spherical steel shell with shell thickness 1.0 mm rollingdown the ramp without slipping.d.A solid aluminum sphere rollingdown the ramp without slipping.

Answer 1

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 = $\frac{1}{2}$ mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  $\frac{1}{2}$ 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 = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ($\frac{2}{5}$mR²) ω²

v = √( $\frac{10}{7}$gh₁  )

so we substitute √( $\frac{10}{7}$gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$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 = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ($\frac{2}{3}$mR²) ω²

v = √( $\frac{6}{5}$gh₁ )

so we substitute √( $\frac{6}{5}$gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( $\frac{6}{5}$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 = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ($\frac{2}{5}$mR²) ω²

v = √( $\frac{10}{7}$gh₁  )

so we substitute √( $\frac{10}{7}$gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$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