Question

Help me please, Two particles each of mass m, are joined by a thin string of length 2L, as shown in Figure 48. A uniform force F is applied in the middle of the string (x = 0) forming a right angle with the initial position of the string. Show that th????

Two particles each of mass m, are joined by a thin string of length 2L, as shown in Figure 48. A uniform force F is applied in the middle of the string (x = 0) forming a right angle with the initial position of the string. Show that the acceleration of each mass in the direction of 90 degrees with F is given `a_x=F/(2m)(x)/(L^2-x^2)^(1/2)`

Answer 1

See below

Explanation:

At all times, where co-ordinate `y` points in the same direction as `F`, Newton's Law for the entire system is:

• `F = 2 m ddot y implies ddot y = F/(2m)`

For each particle, resolving the tension `T` in the string in the direction of, and perpendicular to, motion:

• `T cos alpha = m ddot y`

• `T sin alpha = - m ddot x`

`implies ddot x = - ddot y tan alpha = - F/(2m) tan alpha`

From the annotated drawing:

• `tan alpha = x/(sqrt(L^2 - x^2))`

`implies ddot x = - F/(2m) x/(sqrt(L^2 - x^2)) qquad square`

(NB: There is a minus sign as the `x` direction points left to right away from the axis of symmetry at `x = 0`. The accelerating force is `- T sin alpha \ bb hat x` so that must be correct.)

The Spanish bit then asks about `x = L`, which is right at the start of the motion

Well:

`lim_(x to L) { (ddot x = oo),(T = oo) , (ddot y = 0) :}`

The solution appear to blow up.

But this equation doesn't mean anything at that point in time, as it is derived from there being some angle `alpha ne 0`, etc