Question

An object of mass 0.5 Kg is tied to string and revolved in horizontal circle of radius 1m . If the breaking tension of string is 50N. What is maximum speed of object can have ?

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Answer 1

Answer:

$\boxed{\sf Maximum \ velocity \ (v_{max}) = 10 \ m/s^2}$

Given:

Mass of object (m) = 0.5 kg

Radius of horizontal circle (r) = 1 m

Breaking Tension of string (T) = 50 N

To Find:

Maximum speed ($\sf v_{max}$) that object can have.

Explanation:

Formula:

$\boxed{ \bold{\sf T = \frac{mv_{max}^2}{r}}}$

$\sf \implies v_{max} = \sqrt{ \frac{Tr}{m} }$

Substituting values of T, r & m in the equation:

$\sf \implies v_{max} = \sqrt{ \frac{50 \times 1}{0.5} }$

$\sf \implies v_{max} = \sqrt{ \frac{ \cancel{5} \times 10}{ \cancel{5} \times {10}^{ - 1} } }$

$\sf \implies v_{max} = \sqrt{100}$

$\sf \implies v_{max} = \sqrt{ {10}^{2} }$

$\sf \implies v_{max} = 10 \: m/s$

$\therefore$

Maximum speed ($\sf v_{max}$) that object can have = 10 m/s

Answer 2

This is a simple centripetal force calculation. The tension force required in the string is equal to the object mass times velocity squared divided by the radius of rotation. So first calculate the velocity, then we can convert to a rotational speed.

T = mv^2/r

Rearranging for v:

V = √ Tr/m

V = √ ( 50 × 1 / 0.5)

V = 10 m/s

One revolution circumference is 2π × r = 6.28

So rotational speed is 10/6.28 = 1.59 revolutions per second