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PhyHardAdv 2015 • Paper 2

Tension in Massless Rod Connecting Point Masses Near Fixed Mass

A large spherical mass MM is fixed at one position while two identical point masses mm are positioned along a straight line passing through the centre of MM (refer to the figure). These point masses are joined by a rigid, massless rod of length ll, and the entire assembly can move freely along the line joining them. The three masses interact purely through mutual gravitational attraction. When the point mass closer to MM is located at a distance r=3lr = 3l from MM, the tension in the connecting rod vanishes for m=k(M288)m = k \left( \frac{M}{288} \right). Calculate the value of kk.

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Official Numerical Answer7

Topics & Concepts

GravitationGravitation

Step-by-Step Solution

To find the value of kk, we analyze the forces acting on the two point masses forming the rigid assembly.

Let the mass closer to MM be m1=mm_1 = m at a distance r=3lr = 3l from MM, and the mass farther from MM be m2=mm_2 = m at a distance r+l=4lr + l = 4l from MM.

Let aa be the common acceleration of the two point masses towards the fixed mass MM, and let TT be the tension in the connecting rod.

1. Equation of Motion for the Closer Mass (m1m_1):

The forces acting on m1m_1 along the line connecting the masses are:

  • The gravitational force towards MM: F1M=GMmr2F_{1M} = \frac{GMm}{r^2}
  • The gravitational force away from MM due to m2m_2: F12=Gm2l2F_{12} = \frac{Gm^2}{l^2}
  • The tension TT pulling away from MM

Applying Newton's second law for m1m_1: GMmr2−Gm2l2−T=ma— (1)\frac{GMm}{r^2} - \frac{Gm^2}{l^2} - T = ma \quad \text{--- (1)}

2. Equation of Motion for the Farther Mass (m2m_2):

The forces acting on m2m_2 along the line connecting the masses are:

  • The gravitational force towards MM: F2M=GMm(r+l)2F_{2M} = \frac{GMm}{(r+l)^2}
  • The gravitational force towards MM due to m1m_1: F21=Gm2l2F_{21} = \frac{Gm^2}{l^2}
  • The tension TT pulling towards MM

Applying Newton's second law for m2m_2: GMm(r+l)2+Gm2l2+T=ma— (2)\frac{GMm}{(r+l)^2} + \frac{Gm^2}{l^2} + T = ma \quad \text{--- (2)}

3. Solving for mm when T=0T = 0:

It is given that when r=3lr = 3l, the tension T=0T = 0. Equating expressions (1) and (2) for mama with T=0T = 0:

GMmr2−Gm2l2=GMm(r+l)2+Gm2l2\frac{GMm}{r^2} - \frac{Gm^2}{l^2} = \frac{GMm}{(r+l)^2} + \frac{Gm^2}{l^2}

Dividing both sides by GmG m: Mr2−ml2=M(r+l)2+ml2\frac{M}{r^2} - \frac{m}{l^2} = \frac{M}{(r+l)^2} + \frac{m}{l^2}

Rearranging the terms to isolate mm: 2ml2=M[1r2−1(r+l)2]\frac{2m}{l^2} = M \left[ \frac{1}{r^2} - \frac{1}{(r+l)^2} \right]

Substitute r=3lr = 3l and r+l=4lr+l = 4l: 2ml2=M[1(3l)2−1(4l)2]\frac{2m}{l^2} = M \left[ \frac{1}{(3l)^2} - \frac{1}{(4l)^2} \right]

2ml2=Ml2[19−116]\frac{2m}{l^2} = \frac{M}{l^2} \left[ \frac{1}{9} - \frac{1}{16} \right]

2m=M[16−9144]2m = M \left[ \frac{16 - 9}{144} \right]

2m=M(7144)2m = M \left( \frac{7}{144} \right)

m=7(M288)m = 7 \left( \frac{M}{288} \right)

Comparing this result with m=k(M288)m = k \left( \frac{M}{288} \right), we get: k=7k = 7

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