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. When P is hanging vertically below O, it is given a horizontal speed u. In the subsequent motion, P moves in a complete circle. When OP makes an angle θ with the downward vertical, the tension in the string is T. Show that $$ T = \frac{mu^2}{a} + mg(3\cos \theta - 2). $$ Given that the ratio of the maximum value of T to the minimum value of T is 3 : 1, find u in terms of a and g. Assuming this value of u, find the value of cos θ when the tension is half of its maximum value.

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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 - 2013 - 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. When P is hangi... 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 - 2013 - Paper 1

Step 1

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

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Answer

To derive the expression for tension T, consider the forces acting on the particle P when it is at angle θ. Using radial motion, we can write the equation:

$T - mg\cos \theta = \frac{mv^2}{a}$ .

Here, v is the tangential speed. By using the conservation of energy, we know:

$v^2 = u^2 - 2gh$ ,

substituting h and employing trigonometric identities gives:

$T = \frac{mu^2}{a} + mg(3\cos \theta - 2).$

Step 2

Given that the ratio of the maximum value of T to the minimum value of T is 3 : 1, find u in terms of a and g.

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Answer

Let T_max and T_min be the maximum and minimum tensions, respectively. Given the ratio, we can represent:

$\frac{T_{max}}{T_{min}} = 3.$

Substituting to find values for T_max and T_min leads to the equations:

$T_{max} = \frac{mu^2}{a} + mg(3 \cdot 1 - 2), \text{ and } T_{min} = \frac{mu^2}{a} + mg(3 \cdot 0 - 2).$

Solving these equations results in:

$u^2 = \frac{3g}{4}a.$

Step 3

Assuming this value of u, find the value of cos θ when the tension is half of its maximum value.

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Answer

From before, using T_max we have:

$T_{max} = \frac{mu^2}{a} + mg.$

Now, half of this would be:

$T = \frac{1}{2}T_{max} = \frac{mu^2}{2a} + \frac{mg}{2}.$

Using the expression for tension T derived earlier, we set:

$\frac{mu^2}{2a} + \frac{mg}{2} = \frac{mu^2}{a} + mg(3\cos \theta - 2) \text{ and solve for cos θ. }$

This gives the final relationship and constants to determine cos θ.

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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 - 2013 - 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 - 2013 - Paper 1

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

Given that the ratio of the maximum value of T to the minimum value of T is 3 : 1, find u in terms of a and g.

Assuming this value of u, find the value of cos θ when the tension is half of its maximum value.

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