Question

An object is moving along a circular path of radius 286 cm. If the centripetal force on the object is three times its weight, find the period of motion

Answer 1

k wat is centripetal force with formula?

Answer 2

\[F=(mv ^{2})\div r\]

Answer 3

yeah there the velocity gives u the tangential velocity at that point in circle
so reda ur question and find v
then we will talk abt time period

Answer 4

v = 29.29?

Answer 5

the wat is time period?

Answer 6

im confused :(

Answer 7

61.35?

Answer 8

F=(mv2)÷r
v=2πr/T
F=3B=3mg

Answer 9

time = 6.13 secs?

Answer 10

g=???

Answer 11

what g?

Answer 12

gravitonial acceleration

Answer 13

go step by step relish the feeling u get when u make ur own discovery of the answer

Answer 14

why need to solve for gravitational acceleration?

Answer 15

by using this,

F=(mv2)÷r
\[3m=(mv ^{2}) \div 2.86m\]
v=2.929 m/s

then

T=2πr/v

T=(2π2.86m)/2.929 m/s

T = 6.13 secs

after this?

Answer 16

whats gravitational acceleration?

Answer 17

u have done it that all u got it

Answer 18

thanks

Answer 19

no problem
in the future try following this.....write down the data look for wat u want there will be a slight hint to get what u want to find.....
all the best!

Answer 20

i have another problem, how can i post it?

Answer 21

did u got it cuty?

Answer 22

if centripetal force on the object is three times its weight
that meens F=3*W
W=m*g
so that: 3m=(mv2)÷2.86m is not correct

Answer 23

F=m*V^2/R
3mg=mV^2/R ( bye bye m)
3g=V^2/R
3gR=V^2
V=sqr(3gR)
2πR/T=sqr(3gR)
T=2πR/sqr(3gR)
if g=9.8 m/sec^2 T=1.958=1.96 sec
if g=10 m/sec^2 T=1.939=1.94 sec

Answer 24

3m=(mv2)÷2.86m is not correct

i thought mass = weight .. my bad

Answer 25

so the correct answer is 1.94 secs not 6.13 secs .. ?

Answer 26

whats the correct answer here ..