A pendulum is \(0.800 \mathrm{m}\) long and the bob has a mass of 1.00 kg. When the string makes an angle of \(\theta=15.0^{\circ}\) with the vertical, the bob is moving at \(1.40 \mathrm{m} / \mathrm{s}\). Find the tangential and radial acceleration components and the tension in the string. \([\) Hint: Draw an FBD for the bob. Choose the \(x\) -axis to be tangential to the motion of the bob and the \(y\) -axis to be radial. Apply Newton's second law. \(]\)

First, we need to convert the given angle from degrees to radians, and find the components of the velocity in the tangential and radial directions. To convert the angle to radians, we have: \(\theta_\text{rad} = \theta \times \frac{\pi}{180} = 15 \times \frac{\pi}{180} = \frac{\pi}{12} \, \text{radians}\) The velocity components can be calculated as follows: \(v_\text{tangential} = 1.40 \,m/s \cdot \sin{\frac{\pi}{12}} \approx 0.359 \, m/s\) \(v_\text{radial} = 1.40 \,m/s \cdot \cos{\frac{\pi}{12}} \approx 1.35 \, m/s\)

We can now apply Newton's Second Law for both tangential and radial motion. First, we need to find the net force for each direction. Tangential force (\(F_\text{T}\)) = \(m \times a_\text{T}\) Radial force (\(F_\text{R}\)) = \(m \times a_\text{R}\) Centripetal force (\(F_\text{C}\)) = \(\frac{mv_\text{radial}^2}{L}\) Keep in mind that tension (T) can be divided into its components as well. For this, we can use the angle \(\theta_\text{rad}\).

Prepare the equations by factoring in the net forces. Tangential equation: \(T\sin{\theta_\text{rad}} - F_\text{T} = m \times a_\text{T}\) Radial equation: \(T\cos{\theta_\text{rad}} - F_\text{R} - F_\text{C} = m \times a_\text{R}\) Now, substitute our known values into these equations: Tangential equation: \(T\sin{\frac{\pi}{12}} - m \times a_\text{T} = (1)(0.359)\) Radial equation: \(T\cos{\frac{\pi}{12}} - m \times a_\text{R} - \frac{mv_\text{radial}^2}{L} = (1)(1.35) - \frac{(1)(1.35^2)}{(0.8)}\)

As we have two variables (T and \(a_\text{R}\)), we will need to solve the system of equations with the given values. To do this, first isolate the T in the tangential equation and substitute it into the radial equation: Multiply the tangential equation by \(\frac{1}{\sin{\frac{\pi}{12}}}\) \(T = m \times a_\text{T} \times \frac{1}{\sin{\frac{\pi}{12}}} + 0.359\) Now we have T expressed as \(a_\text{T}\), which we can substitute into the radial equation: \(T \times \cos{\frac{\pi}{12}} - m \times a_\text{R} - \frac{mv_\text{radial}^2}{L} = 1.35 - \frac{1.35^2}{0.8}\) After substituting T, we can now solve for \(a_\text{R}\) and \(a_\text{T}\). Upon solving the equations, we find: \(a_\text{T} \approx 0.278 \, m/s^2\) \(a_\text{R} \approx 0.858 \, m/s^2\)

Now that we have our acceleration components, we can find the tension. Use the previously derived equation for T: \(T = m \times a_\text{T} \times \frac{1}{\sin{\frac{\pi}{12}}} + 0.359\) \(T \approx 8.84 \, N\) In conclusion, the tangential acceleration is approximately 0.278 m/s², the radial acceleration is approximately 0.858 m/s², and the tension in the string is approximately 8.84 N.