Physics Fundamentals
Physics Fundamentals
2nd Edition
ISBN: 9780971313453
Author: Vincent P. Coletta
Publisher: PHYSICS CURRICULUM+INSTRUCT.INC.
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Chapter 5, Problem 35P

(a)

To determine

To Evaluate: The initial location of center of mass for spheres A and B attached to the ends of a rod of negligible mass (Fig. 5-23). The spheres are initially at rest on a frictionless, horizontal surface. A horizontal string is attached to B at point P and is under constant tension of 6.00 N .

  Physics Fundamentals, Chapter 5, Problem 35P , additional homework tip  1

  mA=2.00 Kg

  mB=1.00 Kg

Fig. 5-23

(a)

Expert Solution
Check Mark

Answer to Problem 35P

The initial location of center of mass of the two spheres is 0.333 m

Explanation of Solution

Given:

The mass of spheres A and B are given as mA=2.00 Kg and mB=1.00 Kg respectively and the center to center distance between the spheres is given as 1.00 m . A constant tensional force of T=6.00 N acts on sphere B at point P.

Formula used:

The x-coordinate of center of mass of spheres shown in the figure can be calculated by the following formula:

  xcm=(xAmA+xBmB)(mA+mB)

Where mA and mB are the masses of spheres A and B respectively while xA and xB are the positions of the spheres A and B respectively.

Calculation:

The x-coordinate for the center of mass of spheres is calculated as follows:

  xcm=(xAmA+xBmB)(mA+mB)

Substituting mA=2.00 Kg, xA=0, mB=1.00 Kg and xB=1.00 cm

  xcm=(0×2.00 Kg)+(1.00 m×1.00 Kg)(1.00 Kg + 2.00 Kg)      = 0.333 m

Position of center of mass of the two spheres is thus derived as 0.333 m .

Conclusion:

Thus, the initial location of center of mass of the two spheres is calculated as 0.333 m

(b)

To determine

To Evaluate: The distance by which the center of mass will move in 2.00 s , for spheres A and B attached to the ends of a rod of negligible mass (Fig. 5-23). The spheres are initially at rest on a frictionless, horizontal surface. A horizontal string is attached to B at point P and is under constant tension of 6.00 N .

  Physics Fundamentals, Chapter 5, Problem 35P , additional homework tip  2

  mA=2.00 Kg

  mB=1.00 Kg

Fig. 5-23

(b)

Expert Solution
Check Mark

Answer to Problem 35P

The position of center of mass after 2.00 s will be 4.0 m in the direction of tensional force of 6.00 N .

Explanation of Solution

Given:

The mass of spheres A and B are given as mA=2.00 Kg and mB=1.00 Kg respectively and the center to center distance between the spheres is given as 1.00 m . A constant tensional force of T=6.00 N acts on sphere B at point P.

Formula used: It is known that T=F×r×Sinθ where

F is the force, r is the distance between the point where the force is applied and the axis of rotation and θ is the angle through which the force is applied.

In this case the Cos component of θ has been considered, as the force component along the x-axis needed to be determined.

Calculation:

The net torque that acts on the system is calculated as follows:

  τ=gmA(1.00 m)+gmB(0)

Substituting TCos30(r) for τ and r, the mean distance between the masses is calculated as:

  TCos30=gmA(1.00 m)+gmB(0)           r =gmA(1.00 m)TCos 30

Substituting τ=TCos30, g=9.8 m/s2 and T=6.00 N , and the mean distance r is calculated as:

  r=(9.8 m/ s 2)(2.00 Kg)(1.00 m)(6.00 N)Cos 30   = 3.77 m

This figure could be rounded off to 4.0 m .

Thus, the center of mass gets shifted to 4.0 m in the direction of the tensional force of 6.00 N after 2.00 s .

Conclusion:

Thus, the position of center of mass is shifted by 4.0 m after 2.00 s .

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