How Does Cutting One String Affect the Acceleration of a Rod?

[delecticious]

Homework Statement

A solid rod of mass M = 2.48 kg and length L = 45 cm is suspended by two strings, each with a length d = 35 cm (see Figure), one at each end of the rod. The string on side B is cut. What is the magnitude of the initial tangential acceleration of end B?

The string on side B is retied and now has only half the length of the string on side A. What now is the magnitude of the initial tangential acceleration of end B?

Homework Equations

alpha = I/T

I = r(alpha)

I = 1/3MR^2

The Attempt at a Solution

I'm completely loss really. When string B is cut and it proceeds to swing down would the length of it's string d be counted into the radius? I don't really know what I'm doing here.

 

[PhanthomJay]

Homework Statement

A solid rod of mass M = 2.48 kg and length L = 45 cm is suspended by two strings, each with a length d = 35 cm (see Figure), one at each end of the rod. The string on side B is cut. What is the magnitude of the initial tangential acceleration of end B?

The string on side B is retied and now has only half the length of the string on side A. What now is the magnitude of the initial tangential acceleration of end B?

Homework Equations

alpha = I/T

I = r(alpha)

I = 1/3MR^2

The Attempt at a Solution

I'm completely loss really. When string B is cut and it proceeds to swing down would the length of it's string d be counted into the radius? I don't really know what I'm doing here.

No, don't count the string length in the radius, the rod pivots about its end such that its radius is the same as it length. What you need to find is the initial Torque of the rod's weight from its cg about the pivot. It is different for each case. And correct your second relevant equation to read a_tangential = r(alpha).

 

[ogb p]

I would approach this problem by first identifying the key variables and equations involved. In this case, the key variables are the mass of the rod (M), its length (L), and the lengths of the strings (d). The key equation is the rotational acceleration equation, alpha = I/T, where I is the moment of inertia and T is the torque.

In the first scenario, when the string on side B is cut, the rod will begin to rotate around the string on side A. The moment of inertia (I) for a rod rotating around its center of mass is 1/12ML^2. Using this value for I and the given values for M and L, we can calculate the torque T = Md/2. Plugging these values into the rotational acceleration equation, we can solve for alpha, which will give us the magnitude of the initial tangential acceleration of end B.

In the second scenario, when the string on side B is retied and has half the length of the string on side A, the moment of inertia will change. Since the length of the string is now half, the moment of inertia will be 1/12M(0.5L)^2 = 1/48ML^2. Using this new value for I and the same torque T, we can solve for alpha again and find the new magnitude of the initial tangential acceleration of end B.

In conclusion, the magnitude of the initial tangential acceleration of end B will change in the second scenario due to the change in the moment of inertia. It is important to carefully consider all the variables and equations involved in order to accurately solve the problem.