Circular Motion

Conical pendulum: a small bob of mass $m$ attached to a string of length $L$ fixed at the top, and made to move in a horizontal circle so that the string traces out a cone about the vertical axis through the point of suspension.

Derivation: Let the string make angle $\theta$ with the vertical and the bob move in a circle of radius $r=L\sin\theta$ with speed $v$. The forces on the bob are the tension $T$ along the string and the weight $mg$ downward.

Vertical: no acceleration, so

$T\cos\theta=mg$

Horizontal: the component of tension towards the centre provides the centripetal force,

$T\sin\theta=\frac{mv^2}{r}$

Dividing: $\tan\theta=\dfrac{v^2}{rg}\;\Rightarrow\;v=\sqrt{rg\tan\theta}$

Time period $=$ circumference $/$ speed:

$T_p=\frac{2\pi r}{v}=2\pi\sqrt{\frac{r}{g\tan\theta}}=2\pi\sqrt{\frac{L\sin\theta}{g\tan\theta}}$

$\boxed{T_p=2\pi\sqrt{\frac{L\cos\theta}{g}}}$

The period depends on $L$, $g$ and $\theta$, not on the mass of the bob. It is the same as $2\pi\sqrt{h/g}$ where $h=L\cos\theta$ is the vertical depth of the bob below the point of suspension.

Given $m=0.2\ \text{kg}$, $r=0.5\ \text{m}$, $\theta=30^\circ$ to the vertical, $g=9.8\ \text{m s}^{-2}$.

The vertical component of the tension balances the weight; the horizontal component gives the centripetal force:

$T\cos\theta=mg,\qquad T\sin\theta=\frac{mv^2}{r}$

Tension:

$T=\frac{mg}{\cos 30^\circ}=\frac{0.2\times 9.8}{0.866}\approx 2.26\ \text{N}$

Speed: dividing the two equations, $\tan\theta=\dfrac{v^2}{rg}$, so

$v=\sqrt{rg\tan\theta}=\sqrt{0.5\times 9.8\times 0.577}=\sqrt{2.83}\approx 1.68\ \text{m s}^{-1}$

Angular velocity is the rate of change of angular displacement of a body moving in a circle: $\omega=\dfrac{\theta}{t}$ (unit rad/s).

Relation with linear velocity. Let a particle move in a circle of radius $r$ and cover an arc length $s$ while the radius sweeps through an angle $\theta$ in time $t$. Then

$s=r\theta$

Dividing by $t$,

$\frac{s}{t}=r\,\frac{\theta}{t}$

so, for small $t$,

$\boxed{v=r\omega}$

The linear velocity is tangential to the circle, and for a given $\omega$ it is larger for points farther from the axis. In vector form, $\vec v=\vec\omega\times\vec r$.

(i) Centripetal force is the net force that acts on a body moving in a circle and is directed along the radius towards the centre of the circle. Its magnitude is $\dfrac{mv^2}{L}$.

(ii) The speed is uniform, so at each point the net force towards the centre equals $\dfrac{mv^2}{L}$.

At the lowest point A, the tension acts upwards (towards the centre) and the weight downwards:

$T_A-mg=\frac{mv^2}{L}\;\Rightarrow\;T_A=mg+\frac{mv^2}{L}$

At the highest point B, both the tension and the weight act downwards (towards the centre):

$T_B+mg=\frac{mv^2}{L}\;\Rightarrow\;T_B=\frac{mv^2}{L}-mg$

The tension is greatest at A and least at B.

(iii) When the string is cut at C, the centripetal force vanishes. The bob leaves along the tangent at C with its velocity at that instant, and thereafter, under gravity alone, it follows a parabolic path (projectile motion), not a circle.

(a) No. The magnitude of the centripetal acceleration $v^2/r$ may be constant for uniform motion, but its direction always points to the centre and keeps changing. Acceleration is a vector, so it is not constant.

(b)(i) Resolve $R$ (at $\theta$ above the horizontal):

• vertical component: $R\sin\theta$ (upwards)
• horizontal component: $R\cos\theta$ (towards the centre of the circle)

There is no vertical acceleration, so $R\sin\theta=W$. The weight is cancelled and the resultant force is the horizontal component: $F=R\cos\theta=\frac{W}{\tan\theta}=W\cot\theta$

(ii) $F$ is the resultant force directed towards the centre of the horizontal circle. It is the centripetal force that keeps the ball moving in the circular path: $F=\dfrac{mv^2}{r}$.

(c) $\frac{mv^2}{r}=\frac{mg}{\tan\theta}\;\Rightarrow\;v^2=\frac{rg}{\tan\theta}=\frac{0.14\times9.81}{\tan28^\circ}=\frac{1.373}{0.532}=2.58$ $v\approx1.6\ \text{m s}^{-1}$

(d) work done is zero

The centripetal force is always perpendicular to the velocity, so the work done by it is zero and the kinetic energy stays constant. The body does accelerate (centripetal acceleration), a force acts on it, and the velocity direction changes.

(a) The speed is $v=10\ \text{m s}^{-1}$ at every point, but the direction changes. Change in velocity $\Delta\vec v=\vec v_B-\vec v_A$.

(i) At opposite ends of a diameter the velocities are equal in size and opposite in direction: $|\Delta\vec v|=v-(-v)=2v=20\ \text{m s}^{-1}$

(ii) For $\angle AOB=90^\circ$ the velocity vectors are also $90^\circ$ apart: $|\Delta\vec v|=\sqrt{v^2+v^2}=\sqrt2\,v\approx14.1\ \text{m s}^{-1}$

(b) The acceleration is centripetal (towards the centre) and the velocity is tangential, so the angle between them is $90^\circ$.

(a) Centripetal force is the net force directed towards the centre of a circle that keeps a body moving along a circular path: $F=\dfrac{mv^2}{r}$.

(b) At the highest point, tension $T$ and weight $mg$ both act downward, towards the centre:

$T+mg=\frac{mv^2}{r}$

$T=\frac{0.1\times5^2}{1.0}-0.1\times9.8=2.5-0.98=1.52\ \text{N}$

(c) The particle moves along the tangent to the circle at the point of breaking. At the top this tangent is horizontal, so it initially moves horizontally with speed 5 m/s and then follows a parabolic path under gravity.

(i) Velocity is a vector, with both magnitude and direction. The car's speed is constant, but its direction changes continuously as it goes round the track, so its velocity changes. A changing velocity means acceleration, directed towards the centre of the circle.

(ii) The centripetal force needed is

$F=\frac{mv^2}{r}=\frac{0.50\times(6.0)^2}{1.8}=\frac{18}{1.8}=10\ \text{N}$

The maximum radial friction is only $7.0\ \text{N}$, and $10\ \text{N}>7.0\ \text{N}$. Friction cannot supply the required centripetal force, so the car cannot continue in a circular path.

(iii) At the top of the raised section, the weight $mg$ acts downwards (towards the centre) and the normal reaction $N$ acts upwards. The resultant provides the centripetal force:

$mg-N=\frac{mv^2}{r}$

The car just loses contact when $N=0$:

$mg=\frac{mv_{\max}^2}{r}\ \Rightarrow\ v_{\max}^2=rg\ \Rightarrow\ v_{\max}=\sqrt{rg}$

(i) The car is accelerating because its direction of motion is changing continuously, even though its speed is constant. Velocity is a vector quantity (it has both magnitude and direction), and acceleration is the rate of change of velocity. As the car moves around the circular track, the direction of its velocity at each instant is along the tangent to the circle, and this tangent direction keeps changing from point to point. Since the velocity vector changes (in direction), there is a change in velocity, and hence the car is accelerating. This acceleration is directed towards the centre of the circle (centripetal acceleration), with magnitude $a = \dfrac{v^2}{r}$.

The track radius is $r=1.8\ \text{m}$. The centripetal force needed at $v=6.0\ \text{m s}^{-1}$ is

$F=\frac{mv^2}{r}=\frac{0.50\times(6.0)^2}{1.8}=\frac{18}{1.8}=10\ \text{N}$

The maximum radial friction available is $7.0\ \text{N}$. Since $10\ \text{N}>7.0\ \text{N}$, friction cannot provide enough centripetal force, so the car cannot continue to travel in a circular path.

Derivation of $v_{\max} = \sqrt{rg}$

Consider the car at the highest point of the raised circular section, where the track is curved with radius $r = 0.85\ \text{m}$.

At the highest point, the forces acting on the car are:

• its weight $mg$, acting vertically downwards
• the normal reaction $N$ from the track, acting vertically upwards

These two forces together provide the centripetal force required for circular motion. Taking the downward direction (towards the centre of the circle) as positive:

$mg - N = \frac{mv^2}{r}$

The car loses contact with the track when the normal reaction falls to zero:

$N = 0$

Substituting $N = 0$ into the equation:

$mg - 0 = \frac{mv_{\max}^2}{r}$

$mg = \frac{mv_{\max}^2}{r}$

The mass $m$ cancels from both sides:

$g = \frac{v_{\max}^2}{r}$

Rearranging for $v_{\max}$:

$v_{\max}^2 = rg$

$\boxed{v_{\max} = \sqrt{rg}}$

This is the required relationship. For $r = 0.85\ \text{m}$ and $g = 9.81\ \text{m s}^{-2}$:

$v_{\max} = \sqrt{0.85 \times 9.81} = \sqrt{8.34} \approx 2.9\ \text{m s}^{-1}$

So the car loses contact with the raised section if its speed exceeds about $2.9\ \text{m s}^{-1}$.

Banking of a road is raising the outer edge of a curved road above the inner edge, so that the road surface is inclined at an angle $\theta$ to the horizontal. The inclination lets a vehicle turn safely at higher speed, and reduces dependence on friction and the risk of skidding or overturning.

Motion on a banked track. Consider a vehicle of mass $m$ moving at speed $v$ on a curve of radius $r$ banked at angle $\theta$.

Without friction: the forces are the weight $mg$ and the normal reaction $N$ perpendicular to the surface. The horizontal component of $N$ supplies the centripetal force and the vertical component balances the weight:

$N\sin\theta=\frac{mv^2}{r},\qquad N\cos\theta=mg$

$\tan\theta=\frac{v^2}{rg}\;\Rightarrow\;v=\sqrt{rg\tan\theta}$

This is the ideal (safe) speed for the angle $\theta$.

With friction (coefficient $\mu$): friction acts down the slope at the maximum speed, giving

$N\cos\theta=mg+\mu N\sin\theta,\qquad N\sin\theta+\mu N\cos\theta=\frac{mv^2}{r}$

$v_{\max}=\sqrt{rg\,\frac{\mu+\tan\theta}{1-\mu\tan\theta}}$

A vehicle can therefore take the curve safely at speeds up to $v_{\max}$.