To solve this problem, we need to analyze the balance of moments (torques) acting on the straw when the second insect lands on top of the first one. The straw is supported at one end on the table, and we need to ensure that the center of mass of the system remains above the table to prevent it from toppling.
Understanding the Setup
We have a straw of length \( \frac{3a}{2} \) and mass \( 2m \). Since one third of its length extends beyond the table, the length of the straw that is on the table is:
- Length on the table = \( \frac{3a}{2} - \frac{1}{3} \times \frac{3a}{2} = \frac{3a}{2} - \frac{a}{2} = a \)
This means that the straw extends \( \frac{a}{2} \) beyond the edge of the table. The center of mass of the straw, which is uniform, will be located at its midpoint, which is at a distance of \( \frac{3a}{4} \) from the inner end (the end on the table) and \( \frac{a}{4} \) from the outer end (the end extending beyond the table).
Calculating the Center of Mass
Next, we need to find the position of the center of mass of the entire system when the first insect (mass \( \frac{m}{2} \)) is on the inner end of the straw. The center of mass of the straw is at:
- Position of straw's center of mass = \( \frac{3a}{4} \) from the inner end
Now, when the first insect lands on the inner end, its position is at \( 0 \) (the inner end). The center of mass of the system (straw + first insect) can be calculated using the formula:
\( x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}
\)
Where:
- \( m_1 = 2m \) (mass of the straw)
- \( x_1 = \frac{3a}{4} \) (position of the straw's center of mass)
- \( m_2 = \frac{m}{2} \) (mass of the first insect)
- \( x_2 = 0 \) (position of the first insect)
Substituting these values, we get:
\( x_{cm} = \frac{(2m)(\frac{3a}{4}) + (\frac{m}{2})(0)}{2m + \frac{m}{2}} = \frac{(2m)(\frac{3a}{4})}{\frac{5m}{2}} = \frac{6a}{5} \)
Adding the Second Insect
Now, let’s consider the second insect with mass \( M \) landing on top of the first insect. The new center of mass of the system (straw + first insect + second insect) will be:
\( x_{cm}' = \frac{(2m)(\frac{3a}{4}) + (\frac{m}{2})(0) + M(0)}{2m + \frac{m}{2} + M}
\)
We need to ensure that this new center of mass does not extend beyond the edge of the table, which is at \( a \). Therefore, we set up the inequality:
\( \frac{(2m)(\frac{3a}{4})}{2m + \frac{m}{2} + M} \leq a
\)
Multiplying both sides by \( (2m + \frac{m}{2} + M) \) gives us:
\( (2m)(\frac{3a}{4}) \leq a(2m + \frac{m}{2} + M)
\)
Expanding this, we have:
\( \frac{3ma}{2} \leq 2ma + \frac{ma}{2} + aM
\)
Rearranging terms leads to:
\( \frac{3ma}{2} - 2ma - \frac{ma}{2} \leq aM
\)
Which simplifies to:
\( -ma \leq aM
\)
Dividing by \( a \) (assuming \( a > 0 \)) gives us:
\( -m \leq M
\)
This means that the maximum mass \( M \) of the second insect that can land on the first insect without toppling the straw is \( m \). Therefore, the largest mass of the second insect that can land on the first one without causing the straw to topple is:
m
