Pearson Edexcel IAL Mathematics M3.4.3 Uniform motion of a particle

Practise uniform circular motion for particles on cones, bowls, floors and strings, using force resolution with ω and radius.

[Maximum number: 14]

Figure 5

A particle P of mass m is attached to one end of a light inextensible string of length $a \sqrt{3}$ . The other end of the string is attached to a fixed point A. The particle P is also attached to one end of a second light inextensible string of length a. The other end of this string is attached to a fixed point B, where B is vertically below A, with A B=a.

The particle P moves in a horizontal circle with centre O, where O is vertically below B.
The particle P moves with constant angular speed $\omega$ , with both strings taut, as shown in Figure 5.

(a)

Show that the upper string makes an angle of $30^{\circ}$ with the downward vertical and the lower string makes an angle of $60^{\circ}$ with the downward vertical.

[ 2 ]

(b)

Show that the tension in the upper string is $\frac{1}{2} m \sqrt{3}\left(2 g-a \omega^{2}\right)$ .

[ 8 ]

$T_A\cos30^\circ+T_B\cos60^\circ=mg$
M1 A1
$T_A\sin30^\circ+T_B\sin60^\circ=mr\omega^2$
M1 A1 A1
$r=a\sin60^\circ=\frac{a\sqrt3}{2}$
B1
Solve for $T_A$ .
dM1
$T_A=\frac12m\sqrt3(2g-a\omega^2)$
A1*

(c)

Show that $\frac{2 g}{3 a}<\omega^{2}<\frac{2 g}{a}$

[ 4 ]

Attempt to obtain one inequality on $\omega^2$ .
M1
Correct inequality.
A1
Attempt to obtain another inequality on $\omega^2$ and use both to obtain answer.
M1
$\frac{2g}{3a}<\omega^2<\frac{2g}{a}$
A1*
M1 correct use of either $T_A>0$ or their $T_B>0$ to obtain one inequality on $\omega^2$ .
A1 correct inequality.
M1 use both $T_A>0$ and their $T_B>0$ to form inequalities.
A1* given answer correctly obtained.

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M3.4.3 - Uniform motion of a particle question 1

Question (a)

Question (b)

Question (c)