A 0.25 kg mass attached to a stirng of lenght 1 m is spun in a horizontal circle. At what speed must it?

Question

travel so that the angle the string makes with the horizontal is 30 degree?

Answer 1

Toughie.

T sin(30) supports weight = mg

T cos(30) provides centripetal force = mv^2/r

Divide top by bottom

tan(30) = gr/v^2

so v^2 = gr/tan(30) solve for v

If you draw a triangle with the forces and angle labelled it is more obvious

Answer 2

The answer is: v = 3.836... m/s.

Resolving vertically, T sin 30 = mg ......(A)

Resolving centripetally, T cos 30 = m v^2 / r ......(B)

where r is the radius of the circle it's moving in, which is

r = [sqrt(3)] / 2 metres ......(C) [ ### See note, below. ]

Dividing (A) by (B), tan 30 = r g / v^2, so that

v^2 = r g / tan 30. ......(D)

But tan 30 = 1 /sqrt(3) ......(E)

Using (C) and (E) in (D), v^2 = 9.81 [sqrt(3) / 2] sqrt(3) m^2/s^2

= (3/2) 9.81 m^2/s^2.

Therefore v = 3.836... m/s.

Live long and prosper.

### Note: 'physicsmasta,' immediately above, assumed that ' r ' was 1 metre, but IT'S NOT! (As easily checked, that's how he got 4.12 m/s.) The string has a LENGTH of 1 metre, but it makes a 30 degree angle with the horizontal. So the HORIZONTAL length of the radius the mass is moving in (which is the relevant value of ' r ') is (cos 30) metres = sqrt(3) / 2 metres. That latter value is what I used in my own calculation, of course.

Answer 3

4.12m/s