# Finding angular velocity

#### arze

a disc of moment of inertia $$\displaystyle 0.1kgm^2$$ about its center and radius 0.2m is released from rest on a plane inclined at 30degrees. calculate the angular velocity after it has rolled 2m down the plane if its mass is 5kg.

#### physicsquest

PHF Helper
FInd the height thru which it has come down. The loss in P.E. will be equal to its K.E. Translational + rotational.

arze

#### arze

then i subtract translational k.e. and sqrt{rotational k.e./ (0.5I)}?

#### Parvez

PHF Hall of Honor
mgh = (1/2)Iw^2 + (1/2)mv^2
mg*2sin30 = (1/2)(mR^2/2)w^2 + (1/2)(Rw)^2
mg = (3/4)m R^2*w^2
mg = (6/4) {(1/2)mR^2*w^2}
mg = (3/2) I w^2, (w is the angular celocity)

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#### arze

mgh = (1/2)Iw^2 + (1/2)mv^2
mg*2sin30 = (1/2)(mR^2/2)w^2 + (1/2)(Rw)^2
mg = (3/4)m R^2*w^2
g = (6/4) {(1/2)mR^2*w^2}
g = (3/2) I w^2, (w is the angular celocity)
so i solve for w? i still have some problems with that

#### Parvez

PHF Hall of Honor
I have corrected my previous posting where I inadvertently dropped m on the LHS of the eq in the last two steps.You can alternatively solve it by following way-
mg*2sin30 = (1/2)Iw^2 + (1/2)mv^2
mg = (1/2)(mr^2/2)w^2 + (1/2) m r^2w^2, cancel m both the side and get

g = (3/4)r^2w^2

arze

#### arze

how do you get 3/4? why is it (mr^2/2)?

#### Parvez

PHF Hall of Honor
moment of inertia of a disk I= (1/2)mr^2
The first term on RHS is (1/4)mr^2w^2 and the second one is (1/2)mr^2w^2, hence the sum of the two contains (3/4)mr^2w^2.

arze