A rope of mass M uniformly distributed along its length L ,can slide without friction on a horizontal table. At instant to=0 the hanging part of the rope has a length b and the rope is held at rest .When it is left the rope begins to slide. At instant t the part of the rope that hangs has a length x and a velocity v. Take the level of the horizontal for the gravitational potential energy.
a)Calculate at instant to=0 the mechanical energy of the system in terms of M ,L ,b and g.
b) Calculate the mechanical energy of the system at instant t in terms of M, L, x, v and g.
c) Find relation between x and v
i need the answer
a) To calculate the mechanical energy of the system at instant t=0, we need to consider both the gravitational potential energy and the kinetic energy.
Since the hanging part of the rope has a length b and is at rest, the potential energy can be calculated as mg(b/2), assuming the center of mass of the hanging part is at b/2.
The rope can be approximated as a straight line, and since it is uniformly distributed, the center of mass of the entire rope is at L/2. So, the potential energy contribution from the rest of the rope can be calculated as mg(L/2 - b).
The total potential energy at t=0 is the sum of these two contributions, so the total potential energy is:
PE(t=0) = mg(b/2) + mg(L/2 - b)
The kinetic energy at t=0 is zero because the rope is at rest, so the total mechanical energy at t=0 is:
ME(t=0) = PE(t=0) + KE(t=0) = mg(b/2) + mg(L/2 - b)
b) At instant t, the mechanical energy of the system will include both gravitational potential energy and kinetic energy.
The potential energy contribution can be calculated as mg(x/2), assuming the center of mass of the hanging part is at x/2.
The kinetic energy contribution can be calculated as (1/2)(M/L)xv^2, where x is the length of the hanging part and v is its velocity.
So, the total mechanical energy at t is:
ME(t) = PE(t) + KE(t) = mg(x/2) + (1/2)(M/L)xv^2
c) To find the relation between x and v, we can equate the mechanical energy at t=0 to the mechanical energy at instant t:
mg(b/2) + mg(L/2 - b) = mg(x/2) + (1/2)(M/L)xv^2
Simplifying this equation, we can cancel out the common factor of mg:
(b/2) + (L/2 - b) = (x/2) + (1/2)(M/L)v^2
Combining like terms and rearranging, we get:
L/2 = (x/2) + (1/2)(M/L)v^2 - b
This equation relates x and v for any given values of M, L, b, and g.