An object of 46.5 N is attached to the free end of a light string wrapped around
ID: 1288950 • Letter: A
Question
An object of 46.5 N is attached to the free end of a light string wrapped around a reel with a radius of 0.250 m and mass of 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center. The suspended object is released 5.60 m above the floor.
(a) Determine the tension in the string, the acceleration of the object, and the speed with which the object hits the floor.
(b) Verify your answer by using the principle of conservation of energy to find the speed with which the object hits the floor.
__________m/s
tension __________ N acceleration __________ m/s2 speed __________ m/sExplanation / Answer
Tension T = mg-ma (m = mass of object)
Torque ? = T*R = (mg - ma)*R = I*? = I*(a/R) --> solve for a
(I = 1/2 mR^2 = 1/2*3*0,25^2 = 0,09375 Nm^2)
a = mg/(I/R^2 + m)
a =46.5/(0,09375/0,25^2 + 4.75)
a = 7.44 m/s^2 = acceleration of the object
-------------
Tension = 4.75(9.81-7.44) = 11.2575N
------------
v = sqrt(2sa) = sqrt(2*5.6*7.44)
v = 10.564 m/s = velocity when hitting the ground