Question

A stone is tied to a string (length = 1.05 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 14.6 % larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

A stone is tied to a string (length = 1.05 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 14.6 % larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.
@Mathematics

Answer 1

Solving even these trivial physics problems takes time out of people's busy lives. It'd be nice if you at least said thanks to everyone who's answered your questions.

Answer 2

lol thanks to u all

Answer 3

these physics questions is giving me headace lol

Answer 4

it aint easy at all

Answer 5

can u tell me pls the ans???aftr tht i m showing the procedure

Answer 6

the answer ain't given

Answer 7

A stone is tied to a string (length = 0.800 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 10.0% larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

* Physics - MathMate, Thursday, November 25, 2010 at 1:01pm

Horizontal circle:
Centripetal force (horiz.)+weight of mass(vert.)
Tension on the string, Th
= √((mv²/2)²+(mg)²)

Vertical circle:
maximum tension is when the mass is at the bottom of the circle when the centripetal force is added to the weight.
Tension in the string, Tv
=mv²/2+mg

Equate Tv/Th=1.1
and solve for v.

Answer 8

this a sample question i got on the internet,but dont know how the solved it

Answer 9

pls use it and see if u can arrive to a final answer for my initial question.thanks

Answer 10

u think about the process right bt ur formula ws littl bt wrong
i m thinkng like so

for vertical axis moment

tention=due to the centriptl force- due to mass
Tv=(mv^2/r)-mg

for horizntl case

the centri. force=mv^2/r Th=mv^2/r
now accrdng to ques100 mv^2/r=114.6{(mv^2)/r-mg} now put r as gvn & g=9.81
u wll gt ur ans. it may be 50.6m/s

Answer 11

is this solution for my 1st question?

Answer 12

and what's the mass?