Question

a simple pendulum is made of a small blob of mass ( m = 3.000 mathrm{~kg} ) attached to the end of an inextensible wire. the angular amplitude of oscillation is ( \theta_{0}=5.730^{circ} ).
use ( g = 9.807 mathrm{~m} / mathrm{s}^{2} ) for the gravitational acceleration.
what is the magnitude ( t ) of the tension in the wire when the blob is directly below its point of support? ( t=quad mathrm{n} )

Explanation:

Step1: Use conservation of mechanical energy

The potential energy at the maximum - angle position \(U = mgl(1 - \cos\theta_0)\) (where \(l\) is the length of the pendulum). At the lowest point, the kinetic energy \(K=\frac{1}{2}mv^{2}\). By conservation of energy \(mgl(1 - \cos\theta_0)=\frac{1}{2}mv^{2}\), so \(v^{2}=2gl(1 - \cos\theta_0)\).

Step2: Apply Newton's second law at the lowest point

At the lowest point, the net force towards the center of the circular path (centripetal force) is \(F_{net}=T - mg\). According to Newton's second law \(F_{net}=\frac{mv^{2}}{l}\). Substitute \(v^{2}=2gl(1 - \cos\theta_0)\) into \(T - mg=\frac{mv^{2}}{l}\), we get \(T - mg = 2mg(1 - \cos\theta_0)\). Then \(T=mg(3 - 2\cos\theta_0)\).

First, convert \(\theta_0 = 5.730^{\circ}\) to radians: \(\theta_0=\frac{5.730\pi}{180}\approx0.1\) rad. Using the small - angle approximation \(\cos\theta\approx1-\frac{\theta^{2}}{2}\) (for \(\theta\) in radians), \(\cos(5.730^{\circ})\approx1-\frac{(0.1)^{2}}{2}= 0.995\).

Substitute \(m = 3.000\space kg\), \(g = 9.807\space m/s^{2}\) and \(\cos\theta_0\approx0.995\) into \(T=mg(3 - 2\cos\theta_0)\):

$$
LATEXBLOCK0
$$

Answer:

\(29.61\)