CAIE A-Level Further Math A2 3.3 Circular motion

[Maximum number: 11]

A particle P of mass m is attached to one end of a light inextensible rod of length 3 a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where O Q=x. The system, of rod and particles, rotates about O in a vertical plane.

At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of $60^{\circ}$ from the vertical, the speed of P is $2 \sqrt{a g}$ , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes.

(a)

Show that the speed of Q when the rod has turned through an angle of $60^{\circ}$ from the vertical is

\frac{2 x}{3 a-x} \sqrt{a g} .

[ 2 ]

Angular speeds of P and Q are equal,
so $\frac{\nu_{Q}}{x}=\frac{\nu_{P}}{3 a-x}$ $v_{Q}=\frac{2 x \sqrt{a g}}{3 a-x}$
Shown convincingly: angular speeds equal stated.
AG

(b)

Find x in terms of a.

[ 5 ]

For $P: T+m g \cos 60^{\circ}=\frac{m \times 4 a g}{3 a-x}$
For $Q: T-m g \cos 60^{\circ}=\frac{m v_{Q}^{2}}{x}$
Eliminate $T:-m g \cos 60^{\circ}+\frac{m \cdot 4 a g}{3 a-x}=m g \cos 60^{\circ}+\frac{m v_{Q}^{2}}{x}$ $\begin{aligned} \frac{m \times 4 a g}{3 a-x}=1+\frac{m x 4 a g}{(3 a-x)^{2}} 4 a(3 a-x)=(3 a-x)^{2}+4 a x, x^{2}+2 a x-3 a^{2}=0 \end{aligned}$
Solve to find x.
Obtain 3-term quadratic equation.

$(x-a)(x+3 a)=0, \quad x=a$

(c)

Find u in terms of a and g.

[ 4 ]

KEs correct.

Energy changes from initial position:

Gain in KE of $P: \frac{1}{2} m\left(4 a g-u^{2}\right)$

Loss in KE of $Q: \frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)$

Loss in GPE of $P=m g(3 a-x)\left(1-\cos 60^{\circ}\right)(=m g a)$

Gain in GPE of $Q=m g x\left(1-\cos 60^{\circ}\right)\left(=\frac{1}{2} m g a\right)$

B1FT

GPEs correct.

$\frac{1}{2} m\left(4 a g-u^{2}\right)-\frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)=-m g x\left(1-\cos 60^{\circ}\right)+m g(3 a-x)\left(1-\cos 60^{\circ}\right)$
Energy equation.

Simplify: $4 a g-\frac{5}{4} u^{2}+a g=a g$ $u^{2}=\frac{16}{5} a g, \quad u=\frac{4}{5} \sqrt{5 a g}$
AEF

[Maximum number: 4]

One end of a light inextensible string of length a is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The string is taut and makes an angle $\theta$ with the downward vertical through O, where $\cos \theta=\frac{2}{3}$ . The particle moves in a horizontal circle with speed v.

Find v in terms of a and g.

$\uparrow T \cos \theta=m g$ $\rightarrow T \sin \theta=\frac{m v^{2}}{a \sin \theta}$
Eliminate T and substitute for $\theta$ $v=\sqrt{\frac{5}{6} a g}$
4

[Maximum number: 11]

A particle P of mass m is attached to one end of a light inextensible rod of length 3 a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where O Q=x. The system, of rod and particles, rotates about O in a vertical plane.

At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of $60^{\circ}$ from the vertical, the speed of P is $2 \sqrt{a g}$ , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes.

(a)

Show that the speed of Q when the rod has turned through an angle of $60^{\circ}$ from the vertical is

\frac{2 x}{3 a-x} \sqrt{a g} .

[ 2 ]

Angular speeds of P and Q are equal,
so $\frac{\nu_{Q}}{x}=\frac{\nu_{P}}{3 a-x}$ $v_{Q}=\frac{2 x \sqrt{a g}}{3 a-x}$
Shown convincingly: angular speeds equal stated.
AG

2

(b)

Find x in terms of a.

[ 5 ]

For $P: T+m g \cos 60^{\circ}=\frac{m \times 4 a g}{3 a-x}$
For $Q: T-m g \cos 60^{\circ}=\frac{m v_{Q}^{2}}{x}$
Eliminate $T:-m g \cos 60^{\circ}+\frac{m \cdot 4 a g}{3 a-x}=m g \cos 60^{\circ}+\frac{m v_{Q}^{2}}{x}$ $\begin{aligned} \frac{m \times 4 a g}{3 a-x}=1+\frac{m x 4 a g}{(3 a-x)^{2}} 4 a(3 a-x)=(3 a-x)^{2}+4 a x, x^{2}+2 a x-3 a^{2}=0 \end{aligned}$
Solve to find x.
Obtain 3-term quadratic equation.

$(x-a)(x+3 a)=0, \quad x=a$
5

(c)

Find u in terms of a and g.

[ 4 ]

KEs correct.

Energy changes from initial position:

Gain in KE of $P: \frac{1}{2} m\left(4 a g-u^{2}\right)$

Loss in KE of $Q: \frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)$

Loss in GPE of $P=m g(3 a-x)\left(1-\cos 60^{\circ}\right)(=m g a)$

Gain in GPE of $Q=m g x\left(1-\cos 60^{\circ}\right)\left(=\frac{1}{2} m g a\right)$

B1FT

GPEs correct.

$\frac{1}{2} m\left(4 a g-u^{2}\right)-\frac{1}{2} m\left(\left(\frac{u}{2}\right)^{2}-v_{Q}^{2}\right)=-m g x\left(1-\cos 60^{\circ}\right)+m g(3 a-x)\left(1-\cos 60^{\circ}\right)$
Energy equation.

Simplify: $4 a g-\frac{5}{4} u^{2}+a g=a g$ $u^{2}=\frac{16}{5} a g, \quad u=\frac{4}{5} \sqrt{5 a g}$
AEF

4

[Maximum number: 2]

A particle P of mass m is attached to one end of a light inextensible rod of length 3 a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where O Q=x. The system, of rod and particles, rotates about O in a vertical plane.

At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of $60^{\circ}$ from the vertical, the speed of P is $2 \sqrt{a g}$ , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes.

Show that the speed of Q when the rod has turned through an angle of $60^{\circ}$ from the vertical is

\frac{2 x}{3 a-x} \sqrt{a g} .

Angular speeds of P and Q are equal,
so $\frac{\nu_{Q}}{x}=\frac{\nu_{P}}{3 a-x}$ $v_{Q}=\frac{2 x \sqrt{a g}}{3 a-x}$
Shown convincingly: angular speeds equal stated.
AG

CAIE A-Level Further Math A2 3.3 Circular Motion Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question 4