Physics 1 Lab 11: Angular Momentum

Defining conditions for angular momentum(L) to be conserved

new torque must be zero

linear mechanics: F, Rotational mechanics:

torque (T)

linear mechanics:m, rotational mechanics:

Inertia (I)

linear mechanics: momentum (p), rotational mechanics:

angular momentum L

moment of Inertia for a solid cylinder or disk w mass M and radius R

Ic= 1/2 MR²

Conservation equation for theoretical final angular velocity

Iiωi=(Ic+Ir)ωf

ωftheo (angular velocity after ring is dropped on cylinder:

= Icωi/(Ic+Ir)

angular momentem in therms of I and ω L=

Iω

linear velocity vi and angular velocity ωi on object moving in radius R vi=

ωiR

linear velocity vi and angular velocity ωi on object moving in radius R ωi=

vi/R

% difference between ωftheo and ωfexp

(ωfexp-ωftheo)/((ωfexp+wftheo)/2) x 100%

why does the ωi get faster when the string is pulled inwards

The radius R decreases (I=1/2 MR²), so inertia decreases, L=Iω needs to be conserved so ωi gets bigger

if we graph Lf vs. Li what should the slope be ?

1, because angular momentum should be conserved

why is when the is F applied along the R in a rotational system the torque is 0 ?

T= r x F so if F is on the same plane as r the cross product is zero because the angle between them is either 180 or 0 degrees

what is the key simplification in the lab in relation to torque ?

ignoring small external torques allows us to assume net torque is zero and angular momentum is conserved

what is the moment of inertia for thick-walled cylindrical ring (inner radius ri and outer radius ro)?

½ m(ro²+ri²)

what two quantities do you need to measure before and after dropping the ring

ωi (cylinder alone) and ωf (cylinder and ring)