Question

A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum cable at this point if the pendulum is 75.1 cm long? Submit Answer Tries 0/12 When the pendulum reaches its highest point, what angle does the cable make with the vertical? Submit Answer Tries 0/12 What is the tension in the pendulum cable when the pendulum reaches its highest point?

Answer 1

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, $v_1=$2.58 m/s

a.r=75.1 cm=$75.1\times 10^{-2}m$=0.751 m

$1cm=10^{-2} m$

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

$T=\frac{mv^2}{r}+mg$

Where $g=9.8 m/s^2$

Substitute the values

$T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8$

$T=8.45 N$

b.When the pendulum reaches its highest point,then

Final velocity, $v_2=0$

According to law of conservation of energy

$mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2$

$gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2$

$h_1=0$

Substitute the values

$9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2$

$3.3282=9.8h_2$

$h_2=\frac{3.3282}{9.8}=0.34 m$

The angle mad  by cable with the vertical=$cos\theta=\frac{0.751-0.34}{0.751}=0.55$

$\theta=cos^{-1}(0.55)=56.6^{\circ}$

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

$T=mgcos\theta$

Substitute the values

$T=0.453\times 9.8cos56.6$

$T=2.4 N$