Question

A ball of mass 0.1kg is suspended by a string 30cm long. Keeping the string taut, the ball describes a horizontal circle of radius 15cm. Find the angular speed

Answer 1

Answer:

Mass of the ball, m = 0.1 kgLength of the string, l = 30 cm = 0.30 m

The radius of the horizontal circle, r = 15 cm = 0.15 m

Let us assume that the string makes an angle of “θ” with the vertical, then from the figure attached below we can write

sin θ = 0.15 / 0.30 = 0.5

∴ θ = 30°

In this case there will be no vertical motion so the component of tension in the upward direction i.e., “T cos θ” will be equal to the force due to weight i.e., “mg”, therefore,

T cos θ = mg

Or, T = mg /cosθ ….. (i)

Also, the centripetal force will be given by the component of tension in the horizontal direction i.e., “T sin θ”,

∴ F = T sin θ = [mg /cosθ] * sin θ = mg tan θ ….(ii) [∵ tanθ = sinθ/cosθ]

The formula for centripetal force is given by,

F = m * ω² * r ….. (iii)

Where ω = angular speed (rad/s)

Equating (ii) & (iii) and substituting the given values, we get

mg tan θ = m * ω² * r

or, ω = √[(gtanθ)/r]

or, ω =  $\sqrt{\frac{9.8 * tan 30 }{0.15}}$

or, ω = 6.14 rad/s

Thus, the angular speed is 6.14 rad/s.