Question

An object with a mass of m = 5.5 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.260 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 6.40 m above the floor. (a) Determine the tension in the string. N (b) Determine the magnitude of the acceleration of the object. m/s2 (c) Determine the speed with which the object hits the floor. m/s (d) 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.)

Answer 1

mass of the object,m=5.5 kg

radius, R=0.26 m

mass of the solid disk,M=3 kg

height, h=6.4 m

a)

by using energy relation,

1/2*I*w^2 + 1/2*m*v^2= m*g*h

1/2*(1/2*M*R^2)*(v/R)^2 + 1/2*m*v^2=m*g*h

1/2*(1/2*M*v^2) + 1/2*m*v^2=m*g*h

1/2*(1/2*3*v^2) + 1/2*5.5*v^2 = 5.5*9.8*6.4

===> v=9.93 m/sec

use,

v^2-u^2=2*g*h

9.93^2-0=2*a*6.4

==> a=7.7 m/sec^2

and

T=m*(g-a)

T=5.5*(9.8-7.7)

T=11.55 N

b)

a=7.7 m/sec^2

c)

v=9.93 m/sec

d)

1/2*I*w^2 + 1/2*m*v^2= m*g*h

1/2*(1/2*M*R^2)*(v/R)^2 + 1/2*m*v^2=m*g*h

1/2*(1/2*M*v^2) + 1/2*m*v^2=m*g*h

1/2*(1/2*3*v^2) + 1/2*5.5*v^2 = 5.5*9.8*6.4

===> v=9.93 m/sec