Alight string 4 meters long is wrapped around a solid cylindrical spool with a radius of 0.075 m and a mass of .5 kg. a 5kg mass is then attached to the 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?

a) The angular acceleration of the spool is  2613.33 radians/s^2

b) The spool will be rotating at 210.63 s^-1.

Explanation:

the moment of inertia of the spool, is given by:

I = (1/2)M R^2

= (0.5)(0.5 kg)(0.075m)^2

= 1.406*10^-3 kg m^2

The mass m on a string applies a torque of

T = m g R

  = 5*9.8*0.075

 = 3.675 N-m

(a) Angular acceleration of the spool is

α = T/I  

  = 3.675/1.406*10^-3  

   = 2613.33 radians/s^2

Therefore, the angular acceleration of the spool is  2613.33 radians/s^2 .

(b) The angle rotated through when the string is fully unwound is  

θ = L/(2 π R)  

The time t to unwind is given by

θ = (1/2)*α*T^2

T^2 = 2θ/α

      = 2( L/(2 π R)  )/(α)

      = 2( 4/(2 π 0.075)  )/(2613.33)

      = 6.496*10^-3

T = 0.081

then;

w = α*T

  = 2613.33*0.081

  = 210.63 S^-1

Therefore, the spool will be rotating at 210.63 s^-1.

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

hope this : )

explanation: