Vertical Circular Motion MCQ - Practice Questions with Answers

• This is an example of non-uniform circular motion.

                A particle of mass m is attached to a light and inextensible string. The other end of the string is fixed at O and

              the particle moves in a vertical circle of radius r is equal to the length of the string as shown in the figure.

• Tension at any point on the vertical loop

                                                                         

                 Consider the particle when it is at the point P and the string makes an angle θ with vertical.

                 Forces acting on the particle are:

                                T = tension in the string along its length,

                     And,  mg = weight of the particle vertically downward.

                             Hence, the net radial force on the particle is

$$
\begin{aligned}
F_r & =T-m g \cos \theta \\
\text { And, } F_r & =\frac{m v^2}{r}
\end{aligned}
$$

Where $\mathrm{r}=$ length of the string

$$
\mathrm{So}_3 \frac{m v^2}{r}=T-m g \cos \theta
$$

Or, Tension at any point on the vertical loop

$$
T=\frac{m v^2}{r}+m g \cos \theta
$$
 

                           Since the speed of the particle decreases with height, 

                           hence, tension is maximum at the bottom, where cos θ = 1 (as θ = 0).

$$
T_{\max }=\frac{m v_{\text {Bottom }}^2}{r}+m g
$$

Similarly,

$$
T_{\min }=\frac{m v_{T o p}^2}{r}-m g
$$
 

• Velocity at any point on the vertical loop-

                                                                         

                           If u is the initial velocity imparted to the body at the lowest point then, the velocity of the body at height h is given by

$$
v=\sqrt{u^2-2 g h}=\sqrt{u^2-2 g r(1-\cos \theta)}
$$

- Velocity at the lowest point (A) for the various conditions in Vertical circular motion.
1. Tension in the string will not be zero at any of the points and the body will continue the circular motion.

$$
u_A>\sqrt{5 g r}
$$

2. Tension at highest point C will be zero and the body will just complete the circle.

$$
u_A=\sqrt{5 g r}
$$

3. A particle will not follow the circular motion. Tension in the string becomes zero somewhere between points B and C whereas velocity remains positive. Particle leaves the circular path and follows a parabolic trajectory

$$
\sqrt{2 g r}<u_A<\sqrt{5 g r}
$$

4. Both velocity and tension in the string become zero between $A$ and $B$ and the particle will oscillate along a semi-circular path.

$$
u_A=\sqrt{2 g r}
$$
 

5. The velocity of the particle becomes zero between $A$ and $B$ but the tension will not be zero and the particle will oscillate about the point $A$.

$$
u_A<\sqrt{2 g r}
$$

- Critical Velocity-

It is the minimum velocity given to the particle at the lowest point to complete the circle.

$$
u_A=\sqrt{5 g r}
$$