English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

A ball swings in a vertical circle at the end of a 1.30 m long rope. When the ball is 37.5° past the lowest point on its way up, its total acceleration is (-22.5 i + 20.2 j) m/s2.

Determine the magnitude of its radial acceleration.
(in m/s2)

Determine the speed and velocity of the ball.


I 'm not sure how to solve this problem... I know the basic rotational motion equations, but this is too complicated for me. What I really need is a vector diagram showing the components of its acceleration at that instant, so if anyone could do that for me (in paint or something), that'd be great. Or just show me how to solve the problem.
Thanks.

2007-10-07 15:09:52 · 2 answers · asked by The Rationalist 2 in Science & Mathematics Physics

2 answers

20.2 = a(centrip y) + g
a(centrip y) = 20.2 - g = 20.2 + 9.8 = 30m/s^2
a(resultant due to rot) = √(22.5^2 + 20.2^2) = 30.237m/s^2
f(centripetal) = mv^2/r
f = ma
a = f/m = v^2/r
v^2 = a/r
v = √(a/r) = √(30.237/1.30) = 4.823m/s at 37.5 degrees above the horizontal.

here's the pic:

http://s236.photobucket.com/albums/ff177/jsardi56/?action=view¤t=centripetalacc10-8-07.jpg

2007-10-08 03:39:58 · answer #1 · answered by jsardi56 7 · 0 0

Newton's 2nd law will give you the translational acceleration. Angular acceleration = translational acceleration / radius. The radius is the radius the door swings through, or just its length (2.4 m). Viola. Note that I'm assuming the sections are arranged in a square (2 sections by 2 sections).

2016-05-18 21:03:29 · answer #2 · answered by ? 3 · 0 0

fedest.com, questions and answers