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Expert-verified Found in: Page 292 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # On a horizontal air track, a glider of mass m carries a G-shaped post. The post supports a small dense sphere, also of mass m, hanging just above the top of the glider on a cord of length L. The glider and sphere are initially at rest with the cord vertical. (Figure P9.57 shows a cart and a sphere similarly connected.) A constant horizontal force of magnitude F is applied to the glider, moving it through displacement x1; then the force is removed. During the time interval when the force is applied, the sphere moves through a displacement with horizontal component x2. (a) Find the horizontal component of the velocity of the centre of mass of the glider–sphere system when the force is removed. (b) After the force is removed, the glider continues to move on the track and the sphere swings back and forth, both without friction. Find an expression for the largest angle the cord makes with the vertical. (a) The horizontal component of velocity of centre of mass of glider-sphere system is $\sqrt{\left(\frac{F}{2m}\right)\left({x}_{1}+{x}_{2}\right)}$.

(b) The largest angle of cord from vertical is ${\mathrm{cos}}^{-1}\left[1-\left(\frac{F}{2mgL}\right)\left({x}_{1}+{x}_{2}\right)\right]$.

See the step by step solution

## Step 1: Identification of given data

The mass of glider and sphere is $m$

The length of cord in glider post is $L$

The magnitude of constant horizontal force is $F$

The displacement of glider due to horizontal force is x1

The displacement of sphere due to horizontal force is x2

## Step 2: Center of Mass

The center of mass of system of particles is the point at which total mass of the system can be assumed to be concentrated. At the center of mass the resultant of all the internal forces is zero.

## Step 3: Determination of horizontal component of velocity of center of mass of glider-sphere system

(a)

The average displacement of glider and sphere is given as:

$x=\left(\frac{{x}_{1}+{x}_{2}}{2}\right)$

The constant horizontal force on the system is given as:

$F=\left(m+m\right){a}_{CM}\phantom{\rule{0ex}{0ex}}{a}_{CM}=\frac{F}{2m}$

The horizontal component of velocity of center of mass of glider-sphere system is given as:

${v}_{CM}^{2}={u}^{2}+2{a}_{CM}x$

Here, $u$ is the initial horizontal component of velocity of center of mass of glider-sphere system and its value is zero because system is at rest initially.

For the calculated values of $\text{x}and{\text{a}}_{cm}$, the above equation becomes-

${v}_{CM}^{2}={\left(0\right)}^{2}+2\left(\frac{F}{2m}\right)\left(\frac{{x}_{1}+{x}_{2}}{2}\right)\phantom{\rule{0ex}{0ex}}{v}_{CM}=\sqrt{\left(\frac{F}{2m}\right)\left({x}_{1}+{x}_{2}\right)}$

Therefore, the horizontal component of velocity of center of mass of glider-sphere system is $\sqrt{\left(\frac{F}{2m}\right)\left({x}_{1}+{x}_{2}\right)}$.

## Step 4: Determination of largest angle of cord from vertical

(b)

The height raised by the sphere at largest angle is given as:

$h=L\left(1-\mathrm{cos}\theta \right)$

Here, $\theta$ is the largest angle of cord from vertical.

The speed of glider-sphere system is given as.

${v}_{CM}^{2}=gh$

For the calculated values of ${v}_{CM}and\text{}h$, the above equation becomes-

${\left(\sqrt{\left(\frac{F}{2m}\right)\left({x}_{1}+{x}_{2}\right)}\right)}^{2}=gL\left(1-\mathrm{cos}\theta \right)\phantom{\rule{0ex}{0ex}}1-\mathrm{cos}\theta =\left(\frac{F}{2mgL}\right)\left({x}_{1}+{x}_{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =1-\left(\frac{F}{2mgL}\right)\left({x}_{1}+{x}_{2}\right)\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}\left[1-\left(\frac{F}{2mgL}\right)\left({x}_{1}+{x}_{2}\right)\right]$

Therefore, the largest angle of cord from vertical is ${\mathrm{cos}}^{-1}\left[1-\left(\frac{F}{2mgL}\right)\left({x}_{1}+{x}_{2}\right)\right]$ . ### Want to see more solutions like these? 