What is the tension before the string is cut What is the ten

What is the tension before the string is cut? What is the tension right after the string is cut? What is the tension after rotating to the new position shown?

Solution

We would assume tension in the left string to be T1 and that in the right to be T2. It needs to realised that right before the string it cut, the body is in equilibrium. hence forces in all directions will balanced. However, the moment, the string is cut, it will start accelerating in the direction shown, but the velocity at that very instant would be zero.

At the second point, the particle will also have some velocity, hence it will have an acceleration tangentially as well as radially.

We would balance the forces to determine the tension for the first two parts, then we will have to find the velocity of the body (using conservation of energy) and then determine the velocity.

For before the string is cut:

Vertical direction: T1cosa + T2 cosi = mg

horizontal direction T1 sina = T2 sini

or T1 = T2 sini / sina, putting in the first equation we get:

T2 [(sini * cosa / sina) + cosi] = mg

That is T2 = mg / [(sini * cosa / sina) + cosi]

For right after the string is cut:

Here, it will start accelerating tangentially but will have no radial acceleration [velocity would be zero]

hence, we balance the force along the string:

T2 = mgcosi

For the second point:

Here, the body will also have a radial acceleration because it will have a velocity. By coservation of energy we have: mgL(cosf - cosi) = 0.5* m * v^2

That is: V = sqrt[2gL(cosf - cosi)]

Now, we balance the force radially to determine the tension as:

T2 = (Mv^2 / L) + Mgcosf

or T2 = Mg[2(cosf - cosi) + cosf]