20 Physics -- Circular Motion

If a body connected with a string length r is applied a velocity of √(3rg) at what angle its velocity will become zero

1 Answers

The body is moving in a circular path of radius r and is connected to a string. The velocity of the body is given as √(3rg) and we need to find the angle at which its velocity becomes zero.

The velocity of a body moving in a circular path of radius r is given by v = ωr, where ω is the angular velocity of the body.

The centripetal force required to keep the body moving in a circular path is given by F = mv^2/r = mω^2r.

In this problem, the body is initially moving with a velocity of √(3rg) and is connected to a string of length r. At some point, the tension in the string will become zero and the body will stop moving.

At this point, the centripetal force required to keep the body moving in a circular path will be equal to the weight of the body, which is given by F = mg.

Equating these two forces, we get:

mω^2r = mg

ω^2 = g/r

ω = √(g/r)

At any point in the circular path, the velocity of the body is given by v = ωr.

When the velocity becomes zero, we have:

v = ωr = 0

So, we need to find the angle at which the velocity becomes zero.

We can use trigonometry to find this angle. Let θ be the angle between the string and the vertical.

The vertical component of the tension in the string is given by T cos(θ), and the horizontal component is given by T sin(θ).

At the point where the velocity becomes zero, the tension in the string is zero, so we have:

T cos(θ) = 0

θ = 90°

Therefore, the angle at which the velocity of the body becomes zero is 90°.

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Here,

Coefficient of statistic friction (c) =0.60

Velocity (v)= 110km/hr =30.5m/s

We know,

F = (mv²)/r

Or, ma =mv²/r [F=ma]

Or, a=v²/r [mass is same]

Or, cg =v²/r [acceleration =coefficient of static friction acceleration due to gravity]

Or, 0.6010 =(30.5)²/r

Or, r= 155.041 m.

Therefore the minimum radius is 155.041m.

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