another circular motion question?

Question

A body of mass 1kg is attached to a string 1m long and revolves in a horizontal circle of radius 40cm, thus making a conical pendulum. Find the tension in the string and the period of revolution of the penulum.

Answer 1

First compute the angle of rotation A

A= arcSin(0.4m/1m)=23.8 degrees

The centripetal force Fc=ma
The acceleration a = (V^2) /R
So Fc=m(V^2)/R
Also Fc=W tan(A)
W- weight of the object
W=mg
now Fc=mg tan(23.8)

Let's get back to
Fc=m(V^2)/R
and we have velocity V= sqrt(Fc R / m)
The period T=V/(2 pi R)
Finlay
T=sqrt(Fc R / m)/(2 pi R)
T=sqrt(mg tan(A) R / m)/(2 pi R)
T=sqrt(g tan(A) R )/(2 pi R)or
T=sqrt(g tan(A)/ R )/(2 pi ) just substitute the numbers
T=sqrt(9.81 tan(23.8)/ 0.4)/(2 x 3.14)
T=0.52 sec

Answer 2

OK.. angle A at apex of con pend= 23.56deg
> TsinA=centripet force=m*w^2*r (w=ang vel)
>TcosA=m*g (hence T)
Divide both equns..
>w^2=g*tanA/r> T=2*pi*sqrt(r/(g*tanA)
since T=2*pi/w
> T=1.92sec
(prev resp had T= v/(2*pi*r) instead, so his T would be inverse of 1.92=0.52sec (easily done!))

Answer 3

(a) geometry of a circle => C = 2?R {R = 0.2 hundred m} U can calculate this :>) (b) for tip %: first calculate w = angular speed in rad/s = (1110 rpm) x (2?/60) = ________ rad/s then multiply w via R to locate tip % = V =________ m/s (c) T = era = ans (a) / ans (b) = C/V

Answer 4

I am still drunk from the first project

Answer 5

Simple..0.52 seconds....primary school stuff.