Question

A spider of mass mm is swinging back and forth at the end of a strand of silk of length LL. During the spider's swing the strand makes a maximum angle of θθ with the vertical. What is the speed of the spider at the low point of its motion, when the strand of silk is vertical? Express your answer in terms of the variables mmm, θθtheta, LLL, and ggg.

Answer 1

Answer:

The speed of the spider is v = (2g*L*(1-cosθ))^1/2

Explanation:

using the energy conservation equation we have to:

Ek1 + Ep1 = Ek2 + Ep2

where

Ek1 = kinetic energy = 0

Ep1 = potential energy = m*g*L*cosθ

Ek2 = (m*v^2)/2

Ep2 = m*g*L

Replacing, we have:

0 - m*g*L*cosθ = (m*v^2)/2 - m*g*L

(m*v^2)/2 = m*g*L*(1-cosθ)

v^2 = 2g*L*(1-cosθ)

v = (2g*L*(1-cosθ))^1/2

Answer 2

The speed of the spider is $v =\sqrt{2g*L*(1-cos\theta)}$

Energy conservation:

Energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

From the energy conservation equation we have to:

Ek1 + Ep1 = Ek2 + Ep2

where

Ek1 = kinetic energy = 0

Ep1 = potential energy = m*g*L*cosθ

Ek2 = (m*v²)/2

Ep2 = m*g*L

On substituting the values:

$0 - m*g*L*cos\theta = (m*v^2)/2 - m*g*L\\\\(m*v^2)/2 = m*g*L*(1-cos\theta)\\\\v^2 = 2g*L*(1-cos\theta)\\\\v = \sqrt{(2g*L*(1-cos\theta))}$

Find more information about Energy conservation here:

brainly.com/question/12306849