When a perosn throws a meter stivk it is found that the centre of the stick is moving with speed `10 m//s` and left end stick with speed `20 m//s`. Both points move vertically upwards at that moment. Then angular speed the stick is :

Options

A. 20 rad/sec
B. 10 rad/sec
C. 30 rad/sec
D. none of these

Explanation

To solve the problem of finding the angular speed of the stick when thrown, we can follow these steps:

### Step 1: Understand the problem
We have a meter stick, and we know the speeds of two points on it: the center of the stick and the left end. The center of the stick moves with a speed \( V_c = 10 \, \text{m/s} \) and the left end moves with a speed \( V_p = 20 \, \text{m/s} \). We need to find the angular speed \( \omega \) of the stick.

### Step 2: Define the distance from the center to the end of the stick
The length of the stick is 1 meter, so the distance from the center of the stick to either end is:
\[ r = \frac{1}{2} \, \text{m} \]

### Step 3: Use the relationship between linear speed and angular speed
The relationship between the linear speed of a point on a rotating object and its angular speed is given by:
\[ V = V_c + r \omega \]
where \( V \) is the linear speed of the point, \( V_c \) is the speed of the center, \( r \) is the distance from the center to the point, and \( \omega \) is the angular speed.

### Step 4: Set up the equation for the left end of the stick
For the left end of the stick, we can write:
\[ V_p = V_c + r \omega \]
Substituting the known values:
\[ 20 = 10 + \left(\frac{1}{2}\right) \omega \]

### Step 5: Solve for \( \omega \)
Rearranging the equation gives:
\[ 20 - 10 = \frac{1}{2} \omega \]
\[ 10 = \frac{1}{2} \omega \]
Multiplying both sides by 2:
\[ \omega = 20 \, \text{rad/s} \]

### Conclusion
The angular speed of the stick is \( \omega = 20 \, \text{rad/s} \).

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