Conversion of angular momentum derivations
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Begin with Newton's second law of motion
F⃗ =ma⃗ .F→=ma→.
Multiply both sides by a vector cross product with position gives
r⃗ ×F⃗ Definition of torque=r⃗ ×ma⃗ .r→×F→⏟Definition of torque=r→×ma→.
Using the definition of the cross product, the above equation can be equivalently expressed as
|r⃗ ×F⃗ |=rmasinθ.|r→×F→|=rmasinθ.
Notice that since the force and acceleration are parallel we may consider asinθasinθ as the tangential acceleration atat. Finally this can be cast into final form given by
τ=mr2(at/r),τ=mr2(at/r),
where moment of inertia and angular acceleration are given by
I=mr2,α=at/r,
F⃗ =ma⃗ .F→=ma→.
Multiply both sides by a vector cross product with position gives
r⃗ ×F⃗ Definition of torque=r⃗ ×ma⃗ .r→×F→⏟Definition of torque=r→×ma→.
Using the definition of the cross product, the above equation can be equivalently expressed as
|r⃗ ×F⃗ |=rmasinθ.|r→×F→|=rmasinθ.
Notice that since the force and acceleration are parallel we may consider asinθasinθ as the tangential acceleration atat. Finally this can be cast into final form given by
τ=mr2(at/r),τ=mr2(at/r),
where moment of inertia and angular acceleration are given by
I=mr2,α=at/r,
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