Question

a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br />a. tension in the string<br />b. speed of the body in the circle<br />​

Answer 1

a) T = 2.26 N, b) v = 1.68 m / s

Step-by-step explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

sin 30 =
`(T_x)/(T)`

cos 30 =
`(T_y)/(T)`

Tₓ = T sin 30

T_y = T cos 30

Y axis

T_y -W = 0

T cos 30 = mg (1)

X axis

Tₓ = m a

they relate it is centripetal

a = v² / r

we substitute

T sin 30 = m
`(v^2)/(r)` (2)

a) we substitute in 1

T =
`(mg )/(cos 30)`

T =
`( 0.2 \ 9.8)/(cos \ 30)`

T = 2.26 N

b) from equation 2

v² =
`(T \ sin 30 \ r)/(m)`

If we know the length of the string

sin 30 = r / L

r = L sin 30

we substitute

v² =
`( T \ sin 30 \ L \ sin 30)/(m)`

v² =
`(TL \ sin^2 30)/(m)`

For the problem let us take L = 1 m

let's calculate

v =
`\sqrt{ (2.26 \ 1 \ sin^230)/(0.2) }`

v = 1.68 m / s