physics: rotational dynamics?

Question

A light string 4.00 m long is wrapped around a solid cylindrical spool with a radius of 0.075 m and a mass of 0.500 kg. A 5.00 kg mass is then attached to the free end of the string causing the string to unwind from the spool.
A.) what is the angular acceleration of the spool
B.) How fast will the spool be rotating after all of the string has unwound?

Answer 1

I of the spool is (1/2)*m*r^2

The 5 kg mass puts a torque on the spool
torque = 5 kg*g*r

Angular acceleration, α: torque = I*α
So
5 kg*g*r = (1/2)*m*r^2 * α
Solve for α.

B. Rotation speed ω:
ω^2 = ωo^2 + 2*α*Θ
where ωo=0, α is from part B, Θ can be calculated from the length of the string and the circumference of the spool. (Note: Θ is in radians. 2π radians is one full revolution.)

Answer 2

permit the sector fall ( roll) by using a top h P.E. on the precise is comparable to the sum of the ok.E. by means of translation and that by means of rotation on the backside. mgh = 0.5mv^2 + 0.5 I?^2 ........(one million) I = (2/5)mr^2 v = r? for this reason the above equation (one million) turns into mgh = 0.5 mv^2 + (2/10)mv^2 or mgh = (7/10)mv^2 v^2 = (10/7)gh .....(2) Now use, v^2 = u^2 + 2ah the place v is the linear velocity on the backside , u is the preliminary velocity and "a" is the linear acceleration u = 0 and utilising (2) (10/7)gh = 2ah a = (5/7)g