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)#
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1 Answer
May 29, 2018

See below

Explanation:

OP

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