centripetal force?? 0_o (1 Viewer)

[sukiyaki]

okay i dont get this centripetal force and accerleration i dont know i am lost again~~ as usually in physics

i dont get the below two questions ahh help

1. p, Q and R are three 30kg children on a merry go round 1m , 2m, and 3m, from its centre. It is rotating four times each minute . Calculalate the the centripetal force on each child?

+ seems easy but im lost?

2. A disk was rotated at 5.0hz. Figure below [please excuse me crap diagram] shows 3 sucessive positions of a 25g coin on the desk as taken by a stroboscopic camera. Calculate the centripetal force acting on the coin 1.25 m from teh centre of the disk

 

[McLake]

1. p, Q and R are three 30kg children on a merry go round 1m , 2m, and 3m, from its centre. It is rotating four times each minute . Calculalate the the centripetal force on each child?

Fc = (m[v^2]) / r

v = 2(pi)r / T

F1 = 30 * [(2(pr) * 1) / (4 * 60)] ^2 / 1 = 0.02N

F2 = 30 * [(2(pr) * 2) / (4 * 60)] ^2 / 2 = 0.04N

F3 = 30 * [(2(pr) * 3) / (4 * 60)] ^2 / 3 = 0.06N

2. A disk was rotated at 5.0hz. Figure below [please excuse me crap diagram] shows 3 sucessive positions of a 25g coin on the desk as taken by a stroboscopic camera. Calculate the centripetal force acting on the coin 1.25 m from teh centre of the disk

Fc = (m[v^2]) / r

v = 2(pi)r / T

T = 1 / f

Fc = m * [2(pi)r * f]^2 / r

Fc = 0.025 * [2(pi) * 1.25 * 5]^2 / 1.25 = 30.84N


Sorry about confusing brackets, can you follow?
Are these answers right?

 

[sukiyaki]

oo fanx!!! the last one is correct
but the 1st one is wrong?
mayb the book has wrong answers ??

wells it says p=5.27 / Q = 10.5 // R=15.8

 

[wogboy]

F1 = 30 * [(2(pr) * 1) / (4 * 60)] ^2 / 1 = 0.02N
F2 = 30 * [(2(pr) * 2) / (4 * 60)] ^2 / 2 = 0.04N
F3 = 30 * [(2(pr) * 3) / (4 * 60)] ^2 / 3 = 0.06N

Hmm I believe the mistake is substituting T = 60 * 4, it's supposed to be T = 60/4 = 15. So really, it should look like this:

F1 = 30 * [(2(pr) * 1) / (15)] ^2 / 1 = 5.27N
F2 = 30 * [(2(pr) * 2) / (15)] ^2 / 2 = 10.5N
F3 = 30 * [(2(pr) * 3) / (15)] ^2 / 3 = 15.8N

 

[McLake]

Originally posted by wogboy


Hmm I believe the mistake is substituting T = 60 * 4, it's supposed to be T = 60/4 = 15. So really, it should look like this:

F1 = 30 * [(2(pr) * 1) / (15)] ^2 / 1 = 5.27N
F2 = 30 * [(2(pr) * 2) / (15)] ^2 / 2 = 10.5N
F3 = 30 * [(2(pr) * 3) / (15)] ^2 / 3 = 15.8N

Opps! Thanks for fixing that wogboy ...

 

[underthesun]

not helping you, but,

http://www.geocities.com/nohuntersweb/bigversiongravity2.html

should help you on planetary movements

 

[McLake]

Originally posted by underthesun
not helping you, but,

http://www.geocities.com/nohuntersweb/bigversiongravity2.html

should help you on planetary movements

Have you made any other cool sites besides this one and the "balisitics" one?