Chapter 9: Q12P (page 247)

Two skaters, one with mass $65 kg$ and the other with mass $40 kg$ , stand on an ice rink holding a pole of length $10 m$ and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the $40 kg$ skater move?

Short Answer

Expert verified

The distance covered by the skater $m_{2}$ from the center of mass of the system is $x = 6.2 m .$

Step by step solution

Listing the given quantities

The mass of the one skater is $m_{1} = 65 k g .$

The mass of the second skater is $m_{2} = 40 k g .$

The length of the pole is $I = 10 m .$

Understanding the concept of center of mass

We can use the concept of center of mass of the system. When skaters are pulling each other, the center of mass of the system will not move. Therefore, skaters will meet at the center of mass irrespective of pull from both the skaters.

Formula:

${\vec{R}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$

Calculations of the height of the center of mass

The center of mass of the system does not move. The skaters are meeting at the center of mass. Consider the $m_{2}$ meets distance x from the center of mass.

Substitute the values in the equation (i)

$0 = \frac{(65 k g) (10 m - x) + (40 k g) (- x)}{65 k g + 40 k g} (65 k g) (10 m - x) - (40 k g) (x) = 0 (65 k g \times 10 m) - (65 k g \times x) = (40 k g) (x) (65 k g \times 10 m) = (40 k g) (x) + (65 k g \times x) (65 k g \times 10 m) = (40 k g + 65 k g) (x) x = \frac{(65 k g \times 10 m)}{(40 k g + 65 k g)} = 6.2 m$
The distance covered by the skater $m_{2}$ from the center of mass of the system is $x = 6.2 m$