A man stands at the centre of a circular table with his two hands outstretched. The table is set into rotation with an angular speed of 40 rev/min. What is the angular speed when the man folds his hands back and thereby reduces his moment of inertia 2/5 times the initial value? Assume that the table rotates without friction.

Options

A. A) 100 rpm
B. B) 150 rpm
C. C) 50 rpm
D. D) 75 rpm

Explanation

To solve the problem, we will apply the principle of conservation of angular momentum. Here are the steps to find the final angular speed when the man folds his hands:

### Step 1: Understand the Initial Conditions
- The man is standing at the center of a circular table, and the table is rotating with an initial angular speed (ω_i) of 40 revolutions per minute (rev/min).
- The initial moment of inertia of the man with his hands outstretched is denoted as I₀.

**Hint:** Remember that angular speed is typically measured in revolutions per minute (rpm) or radians per second.

### Step 2: Define the Moment of Inertia After Folding Hands
- When the man folds his hands, his moment of inertia reduces to 2/5 of the initial value. Therefore, the new moment of inertia (I_f) can be expressed as:
\[ I_f = \frac{2}{5} I₀ \]

**Hint:** The moment of inertia is a measure of how mass is distributed relative to the axis of rotation.

### Step 3: Apply the Conservation of Angular Momentum
- According to the conservation of angular momentum, the initial angular momentum (L_i) must equal the final angular momentum (L_f):
\[ L_i = L_f \]
- The angular momentum is given by the product of moment of inertia and angular speed:
\[ L_i = I₀ \cdot ω_i \]
\[ L_f = I_f \cdot ω_f \]

**Hint:** Angular momentum is conserved in a closed system where no external torques are acting.

### Step 4: Set Up the Equation
- Substitute the expressions for angular momentum into the conservation equation:
\[ I₀ \cdot ω_i = I_f \cdot ω_f \]
- Substitute \( I_f = \frac{2}{5} I₀ \):
\[ I₀ \cdot ω_i = \left(\frac{2}{5} I₀\right) \cdot ω_f \]

**Hint:** You can cancel \( I₀ \) from both sides of the equation since it is non-zero.

### Step 5: Solve for the Final Angular Speed
- After canceling \( I₀ \):
\[ ω_i = \frac{2}{5} ω_f \]
- Rearranging gives:
\[ ω_f = \frac{5}{2} ω_i \]
- Now substitute \( ω_i = 40 \) rev/min:
\[ ω_f = \frac{5}{2} \cdot 40 = 100 \text{ rev/min} \]

**Hint:** Ensure that you keep track of units throughout the calculation.

### Final Answer
The final angular speed when the man folds his hands is **100 revolutions per minute (rev/min)**.

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