Question

A 0.400 kg pendulum bob passes through the lowest part of its path at a speed of 3.50 m/s. (a) What is the tension in the pendulum cable at this point if the pendulum is 80.0 cm long? N (b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? ° (c) What is the tension in the pendulum cable when the pendulum reaches its highest point?

Answer 1

(a) the tension in string is

$T=w+F_{c}$

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

$T=0.400*9.8+\frac{0.400*3.50^{2}}{0.80m}$

T= 10.05N

(b)

(KE + PE) 1 + (KE + PEG) 1 +mgyf zmo? +0 = 0+ mgL (1 - cos (max) 1 mgL (1 - cos (max) 1 - cos @max uz 29L v? - 1- 2gL cos max 0 max -1 COS 1 v? 2gL

$\theta _{max}=arccos(1-\frac{3.5^{2}}{2-9.8*0.8})$

$\theta _{max}=77.4^{0}$

(c)

$T=mgcos(\theta _{max})$

$T=0.400*9.81cos(77.4^{0})$

$T=0.856N$