Question

A mass m hangs at the end of a pendulum of length L, which is released at an angle of 40.0° to the vertical. Find the tension in the pendulum cord when it makes an angle of 20.0° to the vertical. [Hint: Resolve the weight along and perpendicular to the cord.]

Answer 1

In order to calculate the tension in the pendulum cod, let's start by sketching the problem:

At this point, all the mechanical energy is potential energy, given by:

$\begin{gathered} h=L-L\cdot\cos (\theta) \\ h=L(1-\cos (40\degree)) \\ h=L(1-\text{0}.766) \\ h=0.234L \\ E_p=m\cdot g\cdot h \\ E_p=m\cdot9.8\cdot0.234L \\ E_p=2.293mL \end{gathered}$

At position 20°, we will have both kinetic and potential energies:

$\begin{gathered} h=L(1-\cos (20\degree)) \\ h=L(1-0.9397) \\ h=0.0603L \\ E_p=m\cdot9.8\cdot0.0603L \\ E_p=\text{0}.5909mL \\ \\ E_k=2.293mL-\text{0}.5909mL=1.7021mL \end{gathered}$

Now, let's decompose the weight force in the directions along (Wy) and perpendicular (Wx) to the cord:

$\begin{gathered} W_x=W\cdot\sin (20\degree) \\ W_x=m\cdot9.8\cdot0.342=3.35m \\ \\ W_y=W\cdot\cos (20\degree) \\ W_y=m\cdot9.8\cdot0.9397=9.209m \end{gathered}$

The sum of the forces along the cord will be equal to the centripetal force:

$\begin{gathered} \sum ^{}_{}F_y=\frac{mv^2}{r} \\ T-W_y=\frac{mv^2}{L} \\ T-9.209m=\frac{mv^2}{L} \end{gathered}$

From the kinetic energy equation, we have:

$\begin{gathered} E_k=\frac{mv^2}{2}=1.7021mL \\ v^2=3.4042L \end{gathered}$

Using this value of v² in the previous equation, we have:

$\begin{gathered} T-9.209m=\frac{m\cdot3.4042L}{L} \\ T=3.4042m+9.209m \\ T=12.6132m \end{gathered}$