An object with a mass of m = 5.1 kg is attached to the free end of a light strin
ID: 1515931 • Letter: A
Question
An object with a mass of m = 5.1 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.240 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 5.40 m above the floor.
(a) Determine the tension in the string.
(b) Determine the magnitude of the acceleration of the object.
s.1 k 0.240 0.240 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 as shown vertical An object with a mass of m = 5.1 kg is attached to the free end of a light string wrapped around a reel of radius R in the figure below. The suspended object is released from rest 5.40 m above the floor m about the a re below. To of m (a) Determine the tension in the string (b) Determine the magnitude of the acceleration of the object m/s 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.)Explanation / Answer
The net vertical force acting on the hanging mass is
Fy = may
mg-T = ma
5.1 * 9.8- T = 5.10 a
torque
Torque = I alpha
TR= I ( a/R)
a = TR^2/(0.5) MR^2)
49.98 N - T = ( 5.1) ( 0.24)^2/0.5 ( 3)( 0.240)^2 )T
=3.4 T
T = 11.36 N
(b)
accleration of the object
a = TR^2/ 1/2* mR^2
= 7.57 m/s^2
(c)
from the kinematic equation
vf^2 = 0+ 2a( hf-hi)
vf= sqrt 2a( hf-hi)
= sqrt 2 ( -7.57) ( 0-5.4)
=9.04 m/s
(d)
by the conservation of energy
Ki+ Ui = Kf + uf
mgh = 1/2 mv^2 + 1/2 I w^2
2mgh = mv^2 + I ( v^2/R^2)
v = sqrt 2 mgh/( m+ I/R^2)
= sqrt 2 ( 5.10)(9.8) ( 5.4)/5.1 + (0.0864 kg m^2/(0.240 m)^2
=9.04 m/s