A small objective placed on a rotating horizontal turn table just slips when it is placed at a distance 4cm from the axis of rotation. If the angular velocity of the trun-table doubled, the objective slip when its distance from the axis of ratation is.

Options

A. 1 cm
B. 2 cm
C. 4 cm
D. 8 cm

Explanation

To solve the problem step by step, we will analyze the forces acting on the object placed on the rotating turntable and how they change when the angular velocity is doubled.

### Step 1: Understand the Forces Acting on the Object
When the object is placed on the rotating turntable, it experiences a centripetal force that keeps it moving in a circular path. This centripetal force is provided by the frictional force between the object and the surface of the turntable.

### Step 2: Write the Expression for Centripetal Force
The centripetal force \( F_c \) required to keep the object moving in a circle of radius \( r \) with angular velocity \( \omega \) is given by:
\[ F_c = m \omega^2 r \]
where \( m \) is the mass of the object, \( r \) is the distance from the axis of rotation, and \( \omega \) is the angular velocity.

### Step 3: Write the Expression for Limiting Friction
The maximum frictional force \( F_f \) that can act on the object before it slips is given by:
\[ F_f = \mu N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a horizontal turntable, the normal force \( N \) is equal to the weight of the object:
\[ N = mg \]
Thus, the maximum frictional force becomes:
\[ F_f = \mu mg \]

### Step 4: Set Up the Condition for Slipping
The object will start to slip when the centripetal force equals the maximum frictional force:
\[ m \omega^2 r = \mu mg \]
Cancelling \( m \) from both sides gives:
\[ \omega^2 r = \mu g \]

### Step 5: Analyze the Initial Condition
Initially, the object slips when it is placed at a distance \( r = 4 \, \text{cm} \). Thus, we can write:
\[ \omega^2 \cdot 4 = \mu g \quad \text{(1)} \]

### Step 6: Analyze the Condition When Angular Velocity is Doubled
When the angular velocity is doubled, we have \( \omega' = 2\omega \). We need to find the new distance \( r' \) at which the object will slip:
\[ (2\omega)^2 r' = \mu g \]
This simplifies to:
\[ 4\omega^2 r' = \mu g \quad \text{(2)} \]

### Step 7: Relate the Two Conditions
From equation (1), we know that \( \omega^2 = \frac{\mu g}{4} \). Substituting this into equation (2):
\[ 4 \left(\frac{\mu g}{4}\right) r' = \mu g \]
This simplifies to:
\[ \mu g r' = \mu g \]
Dividing both sides by \( \mu g \) (assuming \( \mu g \neq 0 \)):
\[ r' = 1 \, \text{cm} \]

### Conclusion
The object will slip when its distance from the axis of rotation is \( 1 \, \text{cm} \).

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