# Pendulum Problem

by Quotexon
Tags: challenge, mechanics, pendulum
 P: 11 1. The problem statement, all variables and given/known data A gorilla is swinging on vines. Each vine is 30 meters long and the gorilla catches each vine when it is at rest 15 degrees to the left of the (downwards) vertical, swings on it until it stops at an angle of 15 degrees to the right of vertical, and then grabs the next vine at rest and repeats the process. What is the gorilla's average horizontal speed in m/s? Details and assumptions You may take g to be 9.8 m/s2. You may treat the gorilla as a simple pendulum and use the small angle approximation. 2. Relevant equations ω=$\sqrt{}g/l$ T=2$\pi$$\sqrt{}L/g$ -gsin$\theta$=d$^{}2$s/dt$^{}2$ 3. The attempt at a solution Since the acceleration in the direction of motion is -gsin$\theta$=d$^{}2$s/dt$^{}2$, i consider taking the integral of this to find the velocity and i sweep on the bounds from -pi/12 to pi/12 radians. I can then find the average velocity. My only concern is what they mean by "horizontal velocity". Also, how can omega be used in this scenario? Any help would be appreciated. Also, sorry for the janky fonts. I'm new to the latex option.
 HW Helper Thanks PF Gold P: 4,247 Average horizontal speed is just horizontal distance traveled divided by the time to travel that distance. Can you find the time to swing from 15 degrees on the left to 15 degrees on the right? Can you find the horizontal distance traveled when going from 15 degrees on the left to 15 degrees on the right? No integration is necessary.
 P: 11 Would I do (2*30sin(15))/T where T = 2pi sqrt(l/g) or (2*30sin(15))*ω where ω= sqrt(g/l)? Thanks
HW Helper
Thanks
PF Gold
P: 4,247

## Pendulum Problem

 Quote by Quotexon Would I do (2*30sin(15))/T where T = 2pi sqrt(l/g) or (2*30sin(15))*ω where ω= sqrt(g/l)?
Almost. What is the time to swing from the left over to the right? It's not T.
 P: 11 Would it be T/2, since T represents the time for a complete cycle? hmm, but would ω have any relevance to the problem? Since it's units are s^-1, shouldn't it be equivalent to simply multiply the distance by the angular frequency?
 HW Helper Thanks PF Gold P: 4,247 Right, you want to use the time of half of a cycle. So, the time is T/2. You don't need to use ω. You can write T in terms of ω as T = 2π/ω, but there's no need to do that here.
 P: 11 Thanks very much, it makes perfect sense at this point! appreciate it!

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