Question

2. [2pt] A mass m = 9.100 kg is suspended from a string of length L = 1.210 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.089 m/s. What is the angle theta between the string and the vertical (in degrees)? Answer: Submit All Answers

Answer 1

Given is:-

Mass   

m = 9.1kg

Length

L= 1.210m

Tangential speed of the mass is   

V = 3.089m/s

Let the radius of the horizontal circle be r

Now,

Using Newton's second law, the net forces are

on horizontal axis eq-1

mv2 Tsint =

and

on vertical axis eq-2

T cost = mg

now dividing eq-1 by eq-2 we get

sino cosA g.r

by plugging   in above equation we get

JUST = 1

sino cose g(Lsino)

by plugging all the values we get

sine cosé (3.089m/s) (9.8m/s) ((1.210m) sino)

or

(sino) coso (3.089m/s) (9.8m/s) ((1.210m))

or

1 - (cose) cose (3.089m/s) (9.8m/s) ((1.210m))

which gives us

cosf = 0.675564

or

A = cos(0.675564)

which gives us

@ = 47.50204°