An object with a mass of m 46 kg is attached to the free en

An object with a mass of m = 4.6 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.270 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 6.20 m above the floor. Determine the tension in the string. Determine the magnitude of the acceleration of the object. Determine the speed with which the object hits the floor, Verify your answer to part (c) by using the isolated system (energy) model. (Do this on paper. Your instructor may ask you to turn in this work.)

Solution

given

m = 4.6 kg

R = 0.2 m

M = 3 kg

a) Let t is the tension in the string and a is the acceleration of the hanging block.

net force acting on hanging block,

Fnet = m*g - T

m*a = m*g - T

T = m*g - m*a

net torque acting on wheel, Tnet = I*alfa (here alfa is angular accel;eration)

T*R = I*a/R

T*R^2 = 0.5*M*R^2*a

(m*g - m*a) = 0.5*M*a

m*g - m*a = 0.5*M*a

m*g = a*(0.5*M + m)

==> a = m*g/(0.5*M + m)

= 4.6*9.8/(0.5*3 + 4.6)

= 7.4 m/s^2

so, T = m*g - m*a

= 4.6*9.8 - 4.6*7.39

= 11.09 N

b) a = 7.4 m/s^2

c) v = sqrt(2*a*h)

= sqrt(2*7.4*6.2)

= 9.58 m/s

d) loss of potentail energy = gain in kinetic energy

m*g*h = 0.5*m*v^2 + 0.5*I*w^2

m*g*h = 0.5*m*v^2 + 0.5*0.5*M*R*w^2

m*g*h = 0.5*m*v^2 + 0.25*M*(R*w)^2

m*g*h = 0.5*m*v^2 + 0.25*M*v^2

v^2 = m*g*h/(0.5*m + 0.25*M)

v = sqrt(m*g*h/(0.5*m + 0.25*M) )

= sqrt(4.6*9.8*6.2/(0.5*4.6 + 0.25*3))

= 9.58 m/s