Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use `costheta=1-theta^2/2 and sintheta=theta` for small theta`.

Explanation

To find the tension in the string when the bob is at an angle of 0.20 radians with a speed of 1.4 m/s, we can follow these steps:

### Step 1: Identify the forces acting on the bob
The forces acting on the bob are:
- The gravitational force (weight) \( mg \) acting downwards.
- The tension \( T \) in the string acting along the string.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
1. A component along the string: \( mg \cos \theta \)
2. A component perpendicular to the string: \( mg \sin \theta \)

However, for our calculation of tension, we only need the component along the string.

### Step 3: Apply Newton's second law in radial direction
For circular motion, the net force towards the center (which is the tension minus the component of weight along the string) provides the centripetal force:
\[ T - mg \cos \theta = \frac{mv^2}{R} \]
Where:
- \( T \) is the tension in the string.
- \( m \) is the mass of the bob.
- \( v \) is the speed of the bob.
- \( R \) is the length of the string (which is also the radius of the circular path).

### Step 4: Rearranging the equation for tension
Rearranging the equation gives:
\[ T = mg \cos \theta + \frac{mv^2}{R} \]

### Step 5: Substitute known values
We need to substitute the values into the equation:
- Given \( v = 1.4 \, \text{m/s} \)
- \( \theta = 0.20 \, \text{radians} \)
- \( m = 0.1 \, \text{kg} \) (assuming from previous problem)
- \( g = 10 \, \text{m/s}^2 \)
- \( R = 1 \, \text{m} \)

### Step 6: Calculate \( \cos \theta \)
Using the approximation \( \cos \theta \approx 1 - \frac{\theta^2}{2} \):
\[ \cos(0.20) \approx 1 - \frac{(0.20)^2}{2} = 1 - \frac{0.04}{2} = 1 - 0.02 = 0.98 \]

### Step 7: Substitute and calculate tension
Now we substitute everything into the tension equation:
\[ T = mg \cos \theta + \frac{mv^2}{R} \]
\[ T = (0.1 \times 10 \times 0.98) + \frac{0.1 \times (1.4)^2}{1} \]
Calculating each term:
1. \( mg \cos \theta = 0.1 \times 10 \times 0.98 = 0.98 \, \text{N} \)
2. \( \frac{mv^2}{R} = \frac{0.1 \times 1.96}{1} = 0.196 \, \text{N} \)

Now, adding these together:
\[ T = 0.98 + 0.196 = 1.176 \, \text{N} \]

### Step 8: Final answer
Thus, the tension in the string at that instant is approximately:
\[ T \approx 1.18 \, \text{N} \]