A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle is held with the string taut and horizontal and is then released. When the string is vertical, it comes into contact with a small smooth peg A which is vertically below O and at a distance x ( < a) from O. In the subsequent motion, when AP makes an angle θ with the downward vertical, the tension in the string is T. Show that $$ T = mg \left( 3 \cos \theta + \frac{2x}{a - x} \right) $$ Given that P completes a vertical circle about A, find the least possible value of $ \frac{x}{a} $.

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A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 3 - 2012 - Paper 1

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A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle is... show full transcript

Worked Solution & Example Answer:A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 3 - 2012 - Paper 1

Step 1

Show that $T = mg(3\cos\theta + \frac{2x}{a - x})$

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Answer

To derive this expression, we need to apply the principle of conservation of energy and Newton's second law.

Initial Speed at Vertical Position: When P is at the lowest point (vertical position), the potential energy is converted to kinetic energy. The speed can be given by:

$v^2 = 2ga$

Therefore, at the point A:

$v = \sqrt{2ga}$
Speed at Angle θ: When the particle P is at an angle θ, the conservation of mechanical energy gives us:

$\frac{1}{2}mv^2 = mg(2a - x) \implies v^2 = 2g(2a - x)$
Application of Forces: Using the radial forces acting on P when it’s at angle θ, we set up the equation:

$T - mg\cos\theta = \frac{mv^2}{a - x}$
Substituting for $v^2$ : Substitute $v^2$ from the previous step:

$T - mg\cos\theta = \frac{m(2g(2a - x))}{a - x}$

Rearranging gives:

$T = mg\cos\theta + \frac{2mg(2a - x)}{a - x}$
Simplifying the Expression: Now simplify the expression:

$T = mg\left(3\cos\theta + \frac{2x}{a - x}\right)$

Hence proved.

Step 2

Find the least possible value of $\frac{x}{a}$

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Answer

To find the least possible value of ( \frac{x}{a} ) when T = 0:

Set T = 0: We begin by considering the condition:

$T = 0 \Rightarrow 0 = mg\left(3\cos\theta + \frac{2x}{a - x}\right) \implies 3\cos\theta + \frac{2x}{a - x} = 0$
Solving the Equation: Rearranging leads to:

$3\cos\theta = -\frac{2x}{a - x}$
Finding $\frac{x}{a}$ :

When θ = π, ( \cos\theta = -1 ):

$3(-1) = -\frac{2x}{a - x} \implies -3 = -\frac{2x}{a - x}$

This results in:

$3(a - x) = 2x \implies 3a - 3x = 2x \implies 3a = 5x \implies \frac{x}{a} = \frac{3}{5}$

Therefore, the least possible value of ( \frac{x}{a} ) is ( \frac{3}{5} ).

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A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 3 - 2012 - Paper 1

Worked Solution & Example Answer:A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 3 - 2012 - Paper 1

Show that $T = mg(3\cos\theta + \frac{2x}{a - x})$

Find the least possible value of $\frac{x}{a}$

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