A light inflexible cable is wrapped around a cylinder of mass m1 , radius R , and moment of inertia about the center of mass Ic . The cylinder rotates about its axis without friction. The cable does not slip on the cylinder when set in motion. The free end of the cable is attached to an object of mass m2 . The object is released from rest at a height h above the floor. You may assume that the cable has negligible mass. Let g be the acceleration due to gravity.
(a) Find the acceleration a of the falling object. Express your answer in terms of m2, R, Ic and g (enter m_2 for m2, R for R, I_c for Ic and g for g).
a=
(b) Find the tension T in the cable. Express your answer in terms of m2, R, Icm and g (enter m_2 for m2, R for R, I_c for Ic and g for g).
T=
(c) Find the speed v of the falling object just before it hits the floor. Express your answer in terms of m2, R, Icm, h and g (enter m_2 for m2, R for R, I_c for Ic, h for h and g for g).
v=
To solve this problem, we can use the principles of conservation of energy and angular momentum.
(a) To find the acceleration of the falling object, we need to consider the conservation of energy. The object starts with potential energy m2gh and ends with kinetic energy (1/2)m2v^2, where v is the final velocity. As there is no friction or other external forces at play, the total mechanical energy is conserved:
m2gh = (1/2)m2v^2
Simplifying, we find:
v^2 = 2gh
v = sqrt(2gh)
The acceleration a is related to the final velocity v and the time taken to fall from height h by the equation v = at. Since the object starts from rest, we have v = 0 when t = 0. Therefore, the time taken to fall from height h is t = sqrt(2h/g).
Substituting the expression for v, we find:
a = v/t = sqrt(2gh) / sqrt(2h/g) = sqrt(gh/g) = sqrt(g)
Therefore, the acceleration of the falling object is a = sqrt(g).
(b) To find the tension T in the cable, we can consider the conservation of angular momentum. Since the cable does not slip on the cylinder, the angular momentum L of the system is conserved. Initially, the object is at rest, so its angular momentum is zero. The cylindrical mass has an angular momentum Lc given by Lc = Icω, where ω is the angular velocity.
When the object falls, the cable unwraps from the cylinder, causing it to rotate. The angular momentum of the falling object is given by m2Rv, where R is the radius of the cylinder. Therefore, we have:
Lc = m2Rv
From the conservation of angular momentum, we have Lc = Icω. Rearranging the equation, we find ω = (m2Rv)/Ic.
The tension T in the cable is related to the angular acceleration α of the cylindrical mass by the equation α = T/Ic. The angular acceleration α is related to the linear acceleration a of the falling object and the radius R by α = a/R. Therefore, we have:
T/Ic = a/R
Simplifying, we find:
T = Ic(a/R) = (m2R^2/Ic)(a/R) = m2a
Therefore, the tension in the cable is T = m2a.
(c) To find the speed v of the falling object just before it hits the floor, we can use the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity (which is zero), a is the acceleration, and s is the distance traveled.
In this case, the distance traveled is the height h. We already found the acceleration a to be sqrt(g) in part (a). Therefore, we have:
v^2 = 0^2 + 2(sqrt(g))(h)
v^2 = 2gh
v = sqrt(2gh)
Therefore, the speed of the falling object just before it hits the floor is v = sqrt(2gh).