A meter stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end when it hits the floor assuming that the end at the floor does not slip is `(g=9.8 m//s^(2))`

Options

A. 3.2 m/s
B. 5.4 m/s
C. 7.6 m/s
D. 9.2 m/s

Answer

To find the speed of the other end of the meter stick when it hits the floor, we can follow these steps:

### Step 1: Understand the system
We have a vertical meter stick (length = 1 meter) that is pivoted at one end (point O) and allowed to fall. The other end (point A) will move in a circular arc around point O.

### Step 2: Identify the energy states
Initially, when the stick is vertical, it has potential energy and no kinetic energy. When the stick falls and the end A hits the floor, it has kinetic energy and no potential energy.

### Step 3: Write the conservation of energy equation
Using the principle of conservation of mechanical energy:
\[ U_A + K_A = U_B + K_B \]
Where:
- \( U_A \) = Initial potential energy at position A
- \( K_A \) = Initial kinetic energy at position A
- \( U_B \) = Final potential energy at position B
- \( K_B \) = Final kinetic energy at position B

### Step 4: Calculate initial potential energy
The height of the center of mass of the stick when it is vertical is \( h = \frac{1}{2} \) meter (0.5 m). Thus, the initial potential energy is:
\[ U_A = mgh = mg \cdot 0.5 \]

### Step 5: Calculate final potential energy
When the stick is horizontal, the height of the center of mass is 0, so:
\[ U_B = 0 \]

### Step 6: Calculate initial and final kinetic energy
Initially, the kinetic energy \( K_A \) is 0 because the stick is at rest. When the stick is horizontal, the kinetic energy can be expressed as rotational kinetic energy about point O:
\[ K_B = \frac{1}{2} I \omega^2 \]
Where \( I \) is the moment of inertia of the stick about point O, and \( \omega \) is the angular velocity.

### Step 7: Calculate the moment of inertia
The moment of inertia \( I \) of a meter stick about one end is given by:
\[ I = \frac{1}{3} mL^2 = \frac{1}{3} m(1^2) = \frac{1}{3} m \]

### Step 8: Substitute into the energy equation
Now substituting into the energy conservation equation:
\[ mg \cdot 0.5 + 0 = 0 + \frac{1}{2} \left(\frac{1}{3} m\right) \omega^2 \]

### Step 9: Simplify the equation
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \cdot 0.5 = \frac{1}{6} \omega^2 \]
\[ \omega^2 = 3g \]
\[ \omega = \sqrt{3g} \]

### Step 10: Calculate the speed of the end A
The speed \( V \) of the end A when it hits the floor is given by:
\[ V = R \omega \]
Where \( R = 1 \) meter (the length of the stick).
Thus:
\[ V = 1 \cdot \sqrt{3g} = \sqrt{3 \cdot 9.8} \]

### Step 11: Calculate the numerical value
Calculating \( V \):
\[ V = \sqrt{29.4} \approx 5.42 \, \text{m/s} \]

### Conclusion
The speed of the other end of the meter stick when it hits the floor is approximately \( 5.42 \, \text{m/s} \).

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