Question

Ela A ball of mass 100gm is suspended by a string 40cm long. Keeping the string always taut, the ball describes a horizontal circle of radius 10cm. Find i. the angle made by the string with the vertical ii. the angular speed of the ball. [2+3] h If the hall in above problem was rotated in a vertical circle what would be​

Answer 1

Hi there! :)

In this situation, the tension in the string is responsible for two things:

- Its vertical component balances the weight of the ball

- Its horizontal component results in a centripetal force

Using the angle from the vertical, the tension's COSINE component would be its vertical component and its SINE component the horizontal component.

i.

We can use right-triangle trigonometry to solve.

We know that the string itself is 40 cm long, and the horizontal leg of the string (opposite side from the vertical angle) is 10 cm.

We can use sine to solve.

Recall:

`sin\phi = (O)/(H)`

Plug in the values and solve for the angle made by the string and the vertical.

`sin\phi = (10)/(40) \\\\sin^(-1)((1)/(4)) = \phi\\\\\phi = \boxed{14.478^o}`

ii.

Begin by summing forces in the vertical direction to solve for tension.

`\Sigma F = T_y - F_g`

Since the forces are balanced (there no acceleration in the vertical direction), ∑F = 0.

`0 = T_y - F_g\\\\T_y = F_g`

The force of gravity (Fg) is equivalent to:

`T_y = F_g = mg`

Ty is the cosine component. We can now solve for the tension.

(Using g≈ 9.8 m/s²)

`Tcos(14.478) = mg \\\\Tcos(14.478)= (0.1)(9.8) \\\\T = ((0.1)(9.8))/(cos(14.478)) = 1.012 N`
We can use this value of 'T' to solve for the net force in the horizontal direction, and subsequently, the ball's angular speed.

Let 'Tx' represent the horizontal component of tension (sine).

In the horizontal direction, we only have the force of tension acting on the ball that results in a net centripetal force.

`F_(xnet) = T_x = m\omega^2 r`

`Tsin\phi = m\omega^2 r`

Rearrange the equation to solve for ω.

`\omega^2 = (Tsin\phi)/(mr)\\\\\omega = \sqrt{(Tsin\phi)/(mr)} = \sqrt{(1.012sin(14.478))/((0.1)(0.1)) }= \boxed{5.03 (rad)/(s)}`

**The last question is unfinished, so I am unable to answer it at this time.