Question

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible.
(a) What is the speed of the rock when the string passes through the vertical position? What is the tension in the string
(b) when it makes an angle of 45 with the vertical,
(c) as it passes through the vertical?

Answer 1

Answer:

(a) $v=11.087\,m.s^{-1}$ & $F_T=1.176\,N$

(b) $F_T=0.8316\,N$

(c) $v=11.087\,m.s^{-1}$ & $F_T=1.176\,N$

Explanation:

Given:

mass of rock, $m=0.12kg$

length of string, $l=0.8\,m$

Max angle of swing from the vertical, θ=45 degree

Height of rise of rock from the mean position:

$cos\theta=\frac{h}{l}$

$cos45^{\circ}=\frac{h}{0.8}$

$h=\frac{0.8}{\sqrt{2} } \,m$

As we know that at the highest point pendulum has highest potential energy and zero kinetic energy.

So,

$PE=m.g.h$

$PE=0.12\times 9.8\times \frac{0.8}{\sqrt{2} }\,J$

At the vertical position of the string we have maximum kinetic energy and zero potential energy:

So,

KE=PE

$\frac{1}{2} \times 0.12 \times v^2=0.12\times 9.8\times \frac{0.8}{\sqrt{2} }$

$v=11.087\,m.s^{-1}$

In vertical position the tension is maximum:

$F_T=m.g cos\theta$

where θ=angle of string from the vertical.

$F_T=9.8\times0.12$

$F_T=1.176\,N$

(b)

As already explained in part (a) the Kinetic Energy is zero so velocity is zero at extreme position.

Now, tension at this point

$F_T=m.g cos\theta$

$F_T=9.8\times0.12 cos45^{\circ}$

$F_T=0.8316\,N$

(c)

The same question has been answered in part (a).