A ring of diameter 1m is rotating with an angular momentum of 10 Joule-sec. The necessary tangential force required to increase its angular momentum by 50 % in 1 sec will be (in newtons)

Options

A. 10
B. 5
C. 15
D. 20

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the initial and final angular momentum
The initial angular momentum \( L_i \) is given as 10 Joule-sec. We need to increase this by 50%.

\[ \text{Increase in angular momentum} = 0.5 \times L_i = 0.5 \times 10 = 5 \text{ Joule-sec} \]

Thus, the final angular momentum \( L_f \) will be:

\[ L_f = L_i + \text{Increase} = 10 + 5 = 15 \text{ Joule-sec} \]

### Step 2: Calculate the change in angular momentum
The change in angular momentum \( \Delta L \) is:

\[ \Delta L = L_f - L_i = 15 - 10 = 5 \text{ Joule-sec} \]

### Step 3: Calculate the torque required
Torque \( \tau \) is defined as the change in angular momentum over time:

\[ \tau = \frac{\Delta L}{\Delta t} \]

Given that the time \( \Delta t \) is 1 second, we have:

\[ \tau = \frac{5 \text{ Joule-sec}}{1 \text{ sec}} = 5 \text{ Newton-meters} \]

### Step 4: Relate torque to tangential force
Torque can also be expressed as the product of tangential force \( F \) and the radius \( r \):

\[ \tau = F \cdot r \]

The diameter of the ring is given as 1 meter, so the radius \( r \) is:

\[ r = \frac{1}{2} \text{ meter} = 0.5 \text{ meter} \]

### Step 5: Solve for the tangential force
Substituting the values into the torque equation:

\[ 5 = F \cdot 0.5 \]

Now, solve for \( F \):

\[ F = \frac{5}{0.5} = 10 \text{ Newtons} \]

### Final Answer
The necessary tangential force required to increase the angular momentum by 50% in 1 second is **10 Newtons**.

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