Asked by Anonymous
If a disk and hoop are released simutaneously 1.0m above the ground at a 30 degree incline which one will reach the bottom faster? Show mathematically.
Answers
Answered by
Elena
PE=KE=KE(transl)+KE(rot)
m•g•h=m•v²/2 +I•ω²/2
ω=v/R
disc
I=m•R²/2 => I•ω²/2= m•R²•v²/2•2• R²=m•v²/4 .
m•g•h=m•v²/2 + m•v²/4=0.75 m•v²
v=sqrt(gh/0.75)
hoop
I=m•R² => I•ω²/2= m•R²•v²/2•R²=m•v²/2 .
m•g•h=m•v²/2 + m•v²/2= m•v²
v=sqrt(gh)
v(disc) > v(hoop) ,
if m(disc)=m(hoop) and R(disc)=R(hoop)
m•g•h=m•v²/2 +I•ω²/2
ω=v/R
disc
I=m•R²/2 => I•ω²/2= m•R²•v²/2•2• R²=m•v²/4 .
m•g•h=m•v²/2 + m•v²/4=0.75 m•v²
v=sqrt(gh/0.75)
hoop
I=m•R² => I•ω²/2= m•R²•v²/2•R²=m•v²/2 .
m•g•h=m•v²/2 + m•v²/2= m•v²
v=sqrt(gh)
v(disc) > v(hoop) ,
if m(disc)=m(hoop) and R(disc)=R(hoop)
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