Acylindrical rod of mass m. length l and radius r has two cords wound around it whose ends are a
to the ceiling as shown. the rod is held horizontally with the two cords vertical. when the rod s mecase
cords unwind the rod rotates. find the tension in the cords as they unwind.

T = mg/6

Explanation:

Draw a free body diagram (see attached).  There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.

The center rotates about the point where the cords touch the edge.  Sum the torques about that point:

∑τ = Iα

mgr = (1/2 mr² + mr²) α

mgr = 3/2 mr² α

g = 3/2 r α

α = 2g / (3r)

(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)

Now, sum of the forces in the y direction:

∑F = ma

2T − mg = m (-a)

2T − mg = -ma

Since a = αr:

2T − mg = -mαr

Substituting expression for α:

2T − mg = -m (2g / (3r)) r

2T − mg = -2/3 mg

2T = 1/3 mg

T = 1/6 mg

The tension in each cord is mg/6.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft