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 /&gt;a. tension in the string
b. speed of the body in the circle
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Answer 1

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

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 = $\frac{T_x}{T}$

        cos 30 = $\frac{T_y}{T}$

        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$\frac{v^2}{r}$            (2)

a) we substitute in 1

         T = $\frac{mg }{cos 30}$

         T = $\frac{ 0.2 \ 9.8}{cos \ 30}$

         T = 2.26 N

b) from equation 2

           v² = $\frac{T \ sin 30 \ r}{m}$

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = $\frac{ T \ sin 30 \ L \ sin 30}{m}$

          v² = $\frac{TL \ sin^2 30}{m}$

For the problem let us take L = 1 m

let's calculate

          v = $\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }$

          v = 1.68 m / s