A Ball Inside a Smooth Cone | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

A ball inside a smooth cone

The word means that no friction acts.
The ball has two real forces: weight and the normal contact force from the cone.
The normal contact force acts at 90° to the cone surface.
The surface is 52° from the horizontal, so the normal contact force is 52° from the vertical.

Keep the path geometry separate from the free-body diagram. Only N and weight are real forces; the dashed arrows are parts of N.

In the free-body diagram, N and weight are the only real forces. The dashed arrows are parts of N. They are not extra forces.
The vertical part of the contact force balances the weight.
The horizontal part is the inward resultant (9702/42/F/M/25 Q1(b)) .

N\cos52^\circ=mg

N\sin52^\circ=\dfrac{mv^2}{r}

Worked example

A ball moving inside a smooth cone

A steel ball moves in a horizontal circle of radius 0.15 m. The cone surface is at 52° to the horizontal. Find the speed and angular speed (9702/42/F/M/25 Q1(c)-Q1(d)).

Divide the inward equation by the vertical equation.
$\dfrac{N\sin52^\circ}{N\cos52^\circ}=\dfrac{ma}{mg}$
$a=g\tan52^\circ$
$a=12.6\text{ m s}^{-2}$
Use the acceleration to find the speed.
$v=\sqrt{ar}$
$v=\sqrt{12.6\times0.15}$
$v=1.4\text{ m s}^{-1}$
Use the speed to find the angular speed.
$\omega=\dfrac{v}{r}$
$\omega=\dfrac{1.4}{0.15}$
$\omega=9.3\text{ rad s}^{-1}$

Answer

v=1.4\text{ m s}^{-1},\quad\omega=9.3\text{ rad s}^{-1}

Your turn9702/42/F/M/25 Q1(e)

The speed of the ball increases. Explain why the path radius must also increase.

Show worked answer

The cone angle does not change, so the acceleration stays constant.

a=g\tan52^\circ=\text{constant}

The speed is squared in the acceleration formula. A larger speed therefore needs a larger radius.