Question

A cylinder of mass 5kg and radius 1m is falling under gravity as the thread is unfolding. What is the tension in Newton in string

Answer 1

Answer:

M a = T - M g      linear acceleration of the center of mass where T = tension

I = Icm + M R^2 = 3/2 M R^2   moment of inertia about the edge of the cylinder

                               using the parallel axis theorem  

M g R = 3/2 M R^2 * (a / R)  where a / R is the angular acceleration and

                                             M g R is the torque about the edge of the cylinder

then  g = 3/2 a

and since T = M (a + g)

T = 5 / 3 M g = 5 / 3 * 5 * 9.8 = 81.7 N

Answer 2

The tension in the string is 0 N.

The tension in the string can be calculated as follows:

Let's consider the forces acting on the cylinder as it falls. The cylinder is subjected to two forces: its weight (W) and the tension (T) in the string.

W = m × g (where m is the mass of the cylinder, g is acceleration due to gravity)

The sum of the forces in the vertical direction is equal to the net force acting on the cylinder, which is given by:

W - T = m × a (where a is the acceleration of the cylinder)

Since the cylinder is in free fall, its acceleration is equal to acceleration due to gravity, i.e. a = g.

Therefore,

W - T = m × g

T = W - m × g

T = $5 kg *9.8 m/s^2 - 5 kg *9.8 m/s^2$

T = 0 N

So the tension in the string is 0 N.

This result occurs because the cylinder is in free fall, and the tension in the string is equal and opposite to its weight, so the net force acting on the cylinder is 0.

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