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A thin rod with massM=5.00kg M=is bent in a semicircle of radiusR=0.650m. (Fig. 13-56). (a) What is its gravitational force (both magnitude and direction on a particle with massm=3.0×10-3kgat P, the center of curvature? (b) What would be the force on the particle the rod were a complete circle?

Short Answer

Expert verified
  1. The magnitude of the gravitational force on the particle =1.5×10-12Nand it is directed upwards (along the +Y axis).
  2. The force on the particle if the rod is a complete circledata-custom-editor="chemistry" =0N.

Step by step solution

01

Listing the given quantities

The mass of the rod M=5.00kg.

The mass of the particlem=3.00×10-3kgand is placed at P, the centre of the semicircle.

The radius of the semicircle R=0.650m.

02

Understanding the concept of Newton’s laws

We can use Newton's law of gravitation and the concept of integration to find the gravitational force.

Formula:

F=GMmr2

03

(a) Calculations for magnitude and direction of the gravitational force

To determine the force exerted by the rod, we will have to calculate its mass per unit length since the rod is in the form of a semicircle.

Mass per unit lengthμ=ML, where L = length of wire =Ï€¸é

Now, consider a small section dl of the rod at an angle θ.

The mass of this section isμ»å±ô.

The force exerted by the section dl on the particle at P is given by .

The total force by all such elements will be directed radially. Hence the net force will be sum of the vertical (sin) components of these forces. All the horizontal (cos) components get cancelled in pairs as they will be directed opposite to each other.

Hence, the net force on particle at P due to semicircular wire is

F=∫F'²õ¾±²Ôθ=∫GMμ»å±ô²õ¾±²ÔθR2

From the figure, we havedl=¸é»åθ.

Hence we write

F=∫GMμ¸é»åθ²õ¾±²ÔθR2=GmMÏ€¸é2∫0Ï€²õ¾±²Ôθdθ

On integrating, we get

F=2GmMÏ€¸é2=2×5×3×10-3×6.67×10-113.14×0.6502=1.51×10-12N

This force will be directed upwards along the + Y axis

04

(b) Calculations of the force on the particle if the rod is complete circle

If the rod is now made in the form of a complete circle, the net force acting on the particle at P will be zero. As we have seen in part (a), all the horizontal components get cancelled in pairs. So, all the vertical components will also get cancelled.

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