Question

A 0.726-kg rope 2.00 meters long lies on a floor. You grasp one end of the rope and begin lifting it upward with a constant speed of 0.710 m/s.

Find the position and velocity of the rope’s center of mass from the time you begin lifting the rope to the time the last piece of rope lifts off the floor.

Plot your results.

(Assume the rope occupies negligible volume directly below the point where it is being lifted.)

Answer 1

Given that :

length of the rope, L = 2 m

constant speed, v = 0.71 m/s

The graph should all end at (t = L / v) which means that t = 2.81 sec, at which time the rope will be entirely off the floor and the velocity of its center of mass will be 0.71 m/s upward.

(a) The position of the center of mass of the rope which will be given as :

using an equation,      x = v t                                                                { eq.1 }

inserting the values in above eq.

x = (0.71 m/s) (2.81 s)

x = 1.99 m

(v) The velocity of the center of mass of the rope which will be given as :

using an equation,    v = m_nf x                                                                  { eq.2 }

the fraction of the rope that is above the floor is given by,

m_nf = (vt / L) M ,      where M - total mass of the rope = 0.726 kg

then,   m_nf = [(1.99 m) / (2 m)] (0.726 kg)

m_nf = 0.72 kg

inserting the value of 'm_nf' in eq.2, we get

v = (0.72 kg) (1.99 m)

v = 1.43 m/s

(c) The plot of Y_cm and V_cm as a function of time which are given below as :