A mass of 0.1 kg is rotated in a vertical circle using a string of legth 20 cm. When the string makes an angle `30^(@)` with the vertical, the speed of the mass is `1.5ms^(-1)`. The tangential acceleration of the mass at that instant is

Options

A. `4.9ms^(-2)`
B. `9.8ms^(-2)`
C. `2.45 ms^(-2)`
D. `19.6 ms^(-2)`

Answer

To find the tangential acceleration of a mass rotating in a vertical circle at a given angle, we can follow these steps:

### Step 1: Identify the Given Values
- Mass (m) = 0.1 kg (not directly needed for tangential acceleration)
- Length of the string (r) = 20 cm = 0.2 m
- Angle with the vertical (θ) = 30°
- Speed (v) = 1.5 m/s
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Determine the Tangential Acceleration
The tangential acceleration (a_t) in circular motion is due to the component of gravitational force acting along the direction of motion. This component can be calculated using the sine of the angle:

\[ a_t = g \cdot \sin(\theta) \]

### Step 3: Calculate the Sine of the Angle
For θ = 30°:

\[ \sin(30°) = \frac{1}{2} \]

### Step 4: Substitute the Values
Now, substituting the values into the equation for tangential acceleration:

\[ a_t = g \cdot \sin(30°) = 9.8 \, \text{m/s}^2 \cdot \frac{1}{2} \]

### Step 5: Perform the Calculation
\[ a_t = 9.8 \cdot 0.5 = 4.9 \, \text{m/s}^2 \]

### Conclusion
The tangential acceleration of the mass at that instant is:

\[ \boxed{4.9 \, \text{m/s}^2} \]

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