Vertical Circular Motion

Vertical Circular Motion

Two types of vertical circular motion are commonly observed.

• A controlled vertical circular motion, such as a giant wheel, in which the speed is either kept constant or is not totally controlled by gravity.
• A vertical circular motion controlled only by gravity. Here the necessary energy is supplied initially, mostly at the lowest point, and thereafter the entire kinetics is governed by the gravitational force, with interconversion of kinetic energy and gravitational potential energy. This is the case the board examines.

Case I: Mass Tied to a String

Diagram

Draw a circle of radius $r$ in a vertical plane with centre $O$. Mark $A$ at the top, $B$ at the bottom, and $C$ and $D$ at the two ends of the horizontal diameter. At $A$ draw both $mg$ and $T_A$ pointing downwards (towards $O$). At $B$ draw $T_B$ upwards (towards $O$) and $mg$ downwards. At $C$ draw $T_C$ horizontally towards $O$ and $mg$ vertically downwards, perpendicular to it. Mark the velocities $v_A$, $v_B$, $v_C$ tangentially.

At any position of the bob only two forces act: its weight $mg$, vertically downwards and constant, and the force due to the tension along the string, directed towards the centre, whose magnitude changes periodically with time and location. Since the motion is non-uniform, the resultant of these two forces is directed towards the centre only at the uppermost and the lowermost positions; at every other position part of the resultant is tangential and changes the speed.

Uppermost position A

Both the weight $mg$ and the force due to the tension $T_A$ are directed downwards, i.e. towards the centre, so their resultant is used entirely as the centripetal force:

$$mg + T_A = \dfrac{mv_A^2}{r} \qquad \dots(1)$$

For the minimum possible speed at this point — that is, if the motion is to be realised with the minimum possible energy — $T_A = 0$, giving

$$(v_A)_{min} = \sqrt{rg} \qquad \dots(2)$$

Lowermost position B

The force due to the tension $T_B$ is vertically upwards, towards the centre, and opposite to $mg$; their resultant is the centripetal force:

$$T_B - mg = \dfrac{mv_B^2}{r} \qquad \dots(3)$$

While coming down from the uppermost to the lowermost point the vertical displacement is $2r$ and the motion is governed only by gravity, so the decrease in gravitational potential energy is converted into kinetic energy:

$$mg(2r) = \dfrac{1}{2}mv_B^2 - \dfrac{1}{2}mv_A^2$$

$$\therefore\ v_B^2 - v_A^2 = 4rg \qquad \dots(4)$$

Using $(v_A)_{min}=\sqrt{rg}$ in (4),

$$(v_B)_{min} = \sqrt{5rg}$$

Midway positions C and D (string horizontal)

Here the force due to the tension is the only force towards the centre, as $mg$ is perpendicular to the tension. Thus the tension is the centripetal force used to change the direction of the velocity, while $mg$ is used only to change the speed:

$$T_C = \dfrac{mv_C^2}{r}$$

Coming down from the midway point to the lowermost point the vertical displacement is $r$, so

$$mgr = \dfrac{1}{2}mv_B^2 - \dfrac{1}{2}mv_C^2 \quad\Rightarrow\quad v_B^2 - v_C^2 = 2gr$$

Since $(v_B)_{min}=\sqrt{5rg}$,

$$v_C^2 = 5rg - 2rg \quad\Rightarrow\quad (v_C)_{min} = \sqrt{3rg}$$

and similarly $(v_D)_{min}=\sqrt{3rg}$.

Difference between the extreme tensions

Subtracting (1) from (3),

$$T_B - T_A - 2mg = \dfrac{m}{r}\left(v_B^2 - v_A^2\right) \qquad \dots(5)$$

Using (4), $\dfrac{m}{r}(4rg) = 4mg$, hence

$$T_B - T_A = 6mg$$

The difference between the tensions at the extreme positions depends only on the weight of the object — it is independent of $v$ and of $r$. By the same working, $T_C - T_A = T_D - T_A = 3mg$.

Arbitrary position

At a general position the force due to the tension and the weight are neither along the same line nor perpendicular. The tangential component of the weight changes the speed — decreasing it while going up and increasing it while coming down. Measuring $\theta$ from the lowermost position,

$$T = \dfrac{mv^2}{r} + mg\cos\theta$$

and the vertical rise above the lowest point is $r(1-\cos\theta)$, which fixes $v$ through energy conservation.

Four cautions from the textbook

• $T_A$ can never be exactly zero for a string, else the string would slacken; so in practice $T_B > 6mg$.
• None of the parameters — including the linear and angular accelerations — is constant during this motion, so the kinematical equations of Table 1 do not apply.
• The position or velocity at any instant can be found from energy conservation, but the radius-vector function is not integrable, so the period or frequency cannot be determined theoretically; it can only be measured experimentally.
• In reality energy must be supplied continuously to overcome air resistance.

Case II: Mass Tied to a Rod

The basic difference between a rod and a string is that a string needs some tension at all points, including the uppermost. A rigid rod can push as well as pull, so no such condition is necessary and a (practically) zero speed is possible at the uppermost point. By similar mathematics,

$$(v_{lowermost})_{min}=\sqrt{4rg}=2\sqrt{rg}, \qquad (v_{midway})_{min}=\sqrt{2rg}, \qquad (v_{uppermost})_{min}=0$$

$$T_{lowermost} - T_{uppermost} = 6mg$$

Sphere of Death

Two-wheeler riders undergo rounds inside a hollow sphere, starting with small horizontal circles and eventually performing revolutions along vertical circles. The dynamics of this vertical circular motion is exactly the same as that of a point mass tied to a string, except that the force due to the tension $T$ is replaced by the normal reaction force $N$. In the initial nearly horizontal circles the linear speed is greater for the larger circles while the angular speed (frequency) is greater for the smaller ones, as the theory of the conical pendulum predicts.

Vehicle at the Top of a Convex Over-Bridge

Diagram

Draw the arc of a convex bridge of radius of curvature $r$ with its centre below the road. Place the vehicle at the topmost point and draw $mg$ vertically downwards and $N$ vertically upwards.

At the uppermost position the resultant of the two vertical forces must provide the centripetal force, which points vertically downwards towards the centre:

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

As the speed increases, $N$ goes on decreasing. The normal reaction is an indication of contact, so for contact to be just maintained $N=0$, giving an upper limit

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

Standard Results

• Top of the circle: $mg+T_A=\dfrac{mv_A^2}{r}$, $\ (v_A)_{min}=\sqrt{rg}$
• Midway: $T_C=\dfrac{mv_C^2}{r}$, $\ (v_C)_{min}=\sqrt{3rg}$
• Bottom: $T_B-mg=\dfrac{mv_B^2}{r}$, $\ (v_B)_{min}=\sqrt{5rg}$
• $v_B^2-v_A^2=4rg$, $\quad v_B^2-v_C^2=2rg$
• $T_B-T_A=6mg$, $\quad T_C-T_A=3mg$
• Any position, $\theta$ from the bottom: $T=\dfrac{mv^2}{r}+mg\cos\theta$, rise $=r(1-\cos\theta)$
• Rod: $0$, $\sqrt{2rg}$, $2\sqrt{rg}$ at top, midway and bottom
• Convex over-bridge: $mg-N=\dfrac{mv^2}{r}$, $\ v_{max}=\sqrt{rg}$