Question

Your FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg). He had the only paddle but managed to drop it in the lake. It has drifted just out of his reach along the long axis of the canoe. His sister has taken Physics 121 and suggests that if they trade places she might be able to reach the paddle. Due to the high "slime" content of the lake and the slick surface of the canoe, the canoe moves in the water without any friction or drag. The passengers sit 2.03 meters apart and the canoe has a mass of 48.7 kg. After the passengers switch places, how much closer to the paddle (which does not move) is the end of the canoe? (Enter a positive number.)

Answer 1

Answer:

• The end of the canoe is  46.27 cm closer to the paddle.

Explanation:

This problem can be solved using the center of mass of the system.

As there are no external forces in the horizontal direction, and the external forces in the vertical direction, the gravitational force and the buoyancy,  cancel each other, we know:

$\vec{F}_{net_{external}} = M \frac{d\vec{P}}{dt} = 0$

where M is the total mass of the system, and $\vec{P}$ is the linear momentum of the system. This implies:

$\frac{d\vec{P}}{dt} = 0$

so

$\vec{P} = constant$

the linear momentum of the system is:

$\vec{P} = M \vec{r}_{cm}$

where $\vec{r}_{cm}$ is the position of the center of mass. As the mass in this problem is constant, this implies that the position of the center of mass is conserved.

Finding the center of mass

For a system formed by i particles, each of mass $m_i$ and position $\vec{r}_i$, the center of mass can be found with the equation:

$\vec{r}_{cm} = \frac{\sum\limits_i m_i \vec{r}_i }{M}$

where M is the total mass of the system, found with the equation:

$M = \sum\limits_i m_i$

For this problem, the total mass of the system is :

$M = m_{brother} + m_{sister} + m_{canoe}$

$M = 144.1 \ kg + 81.6 \ kg + 48.7 \ kg$

$M = 274.4 \ kg$

To find the positions, we need an frame of reference.  Putting the origin of coordinate at the position of the sister.

$\vec{r}_{sister_i} = 0 \ \hat{i}$

The brother is 2.03 meters apart in the positive direction:

$\vec{r}_{brother_i} = 2.03 m \ \hat{i}$

and the center of mass of the canoe must be at half this distance:

$\vec{r}_{canoe_i} = 1.015 m \ \hat{i}$

taking all this, the center of mass of the system is at :

$\vec{r}_{cm} = \frac{ 81.6 \ kg * 0 \ \hat{i} + 144.1 \ kg \ 2.03 m \ \hat{i} + 48.7 \ kg \ 1.015 m \ \hat{i} }{274.4 \ kg}$

$\vec{r}_{cm} = 1.2462 \ m\ \hat{i}$

Finding the displacement

After the switch, we found that the center of mass has to be displaced. If its displaced a distance d, we find that the new positions must be:

$\vec{r}_{sister_f} = 2.03 m \ \hat{i} + d \ \hat{i}$

$\vec{r}_{brother_f} = 0 \ \hat{i} + d \ \hat{i}$

$\vec{r}_{canoe_f} = 1.015 m \ \hat{i} + d \ \hat{i}$

where the brother and the sister had switched the original position. Now, the center of mass will be located at:

$\vec{r}_{cm} = 1.2462 \ m\ \hat{i}$

And the equation is :

$\vec{r}_{cm} = \frac{ 81.6 \ kg * (2.03 m \ \hat{i} + d  \ \hat{i} )  + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ ( 1.015 m \ \hat{i} + d \ \hat{i}) }{274.4 \ kg}$

$\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg  d  \ \hat{i} )  + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ d \ \hat{i}) }{274.4 \ kg}$

$\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg   + 144.1 \ kg   + 48.7 \ kg \ }{274.4 \ kg} \ d \ \hat{i}$

$\vec{r}_{cm} = 0.7838 \ m \ \hat{i} +  d \ \hat{i}$

$1.2462 \ m\ \hat{i} = 0.7838 \ m \ \hat{i} +  d \ \hat{i}$

$d \hat{i} = 0.4627 \ m \ \hat{i}$

And this is how much closer the paddle is to the end of the canoe.

Answer 2

After the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.

What is center of mass?

The center of mass of a object or for a system of object is the point, where the center of distribution of mass in space.

FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg) and canoe has a mass of 48.7 kg.

Let the position of the sister at the origin, his brother is at 2.03 meters. The position of the center of the mass of the canoe is at half distance (1.015 m). Thus, the center of mass of the system is,

$r_{COM}=\dfrac{144.1(2.03)+81.6(0)+48.7(1.015)}{144.1+81.6+48.7}\\r_{COM}=1.2462$

For the displacement, the equation can be given as,

$r_{COM}=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\1.2462=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\d=0.4627\rm \; m$

Thus, after the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.

Learn more about the center of mass here;

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