Learn centripetal acceleration & force definition and formula, Practice problems to find minimum speed to cross the top most point of path of string in vertical circular Motion.

- It is a component of net acceleration which is perpendicular to velocity & acting towards the center in circular motion.
- The magnitude of centripetal acceleration is \(\dfrac{v^2}{R}\), where v is the speed of the particle and R is the radius of curvature.

- It is the net force acting towards the radial direction and directed towards center in circular motion.

\(\vec{F}_{C} = \dfrac{mv^2}{R} \)

- Consider a mass m, suspended from ceiling by a light string moving along a horizontal circle, as shown in figure. This arrangement is known as conical pendulum.
- The radius of the circle is R.
- Let the speed of the mass is v.

**\(\sum F_C = T sin\,\theta\)**

By Newton's second law,

\(Tsin\,\theta = ma_C\)

\( \Rightarrow \; T sin\, \theta = \dfrac{mv^2}{R} \) ...(1)

Acceleration = 0 in vertical direction

\(\Rightarrow \; T cos \; \theta - mg = 0\)

\(\Rightarrow \; T cos \; \theta = mg\) ...(2)

- From (1) & (2)

\(\Rightarrow \; tan\, \theta = \dfrac{v^2}{ Rg}\)

\(v = \sqrt {Rg\; tan\,\theta} \)

A \(2 \sqrt 5\; m/s\)

B \(\sqrt10\; m/s\)

C \(5 \; m/s\)

D \(10 \; m/s\)

- To avoid skidding of vehicle at the turn, the roads are banked at some angle.
- For this, the outer part of the road is lifted up, as compared to the inner part, as shown in figure.

\(\sum F_C = N \,sin\,\theta \)

Using Newton's second law,

\(N \, sin\, \theta = \dfrac{mv^2}{R}\) ...(1)

where R is the radius of curvature

v is the speed of vehicle

m is the mass of vehicle

\(N \, cos\, \theta -mg = 0\)

\(N \, cos \,\theta = mg\) ...(2)

- From equation (1) & (2)

\(tan\,\theta = \dfrac{v^2}{Rg}\)

\(\theta = tan^{-1} \left(\dfrac{v^2}{Rg}\right)\)

where \(\theta\) is the angle of banking of road.

- Consider a ball of mass m, attached with a string.
- The ball is moving with speed v in vertical circular motion, as shown in figure.

- Applying Newton's second law along radial direction,

\(mg + T = \dfrac{mv^2}{R}\)

- For minimum speed of the ball to cross the top most point without slacking of string, the tension force becomes very low.
- The condition just before slacking of string is

\(mg \leq \dfrac{mv^2}{R}\)

\(v_{min} = \sqrt {Rg}\)

**Important Note**

In this situation, neither the tension force is zero nor the string slacks. Here, minimum value of speed is calculated for just before the slacking of string.

A \(\sqrt 2 \; m/s\)

B \(\sqrt 3\; m/s\)

C \(2 \; m/s\)

D \(3 \; m/s\)

- Consider a ball of mass m, attached with two strings, as shown in figure.
- The system is in circular motion about a rod P with constant velocity v in a circle of radius R.

\(T_1 \; cos \,\theta _1 + T_2\; cos\, \theta _2 = \sum\; F_C\)

Using Newton's second law,

\(T_1 \; cos \,\theta _1 + T_2 \; cos\, \theta_2 = \dfrac{mv^2}{R}\) ...(1)

\(T_1 \; sin\, \theta _1 - T_2 \; sin\, \theta _2 - mg = 0\)

\(T_1 \; sin\, \theta _1 - T_2 \; sin\, \theta _2 = mg \) ...(2)

- On solving equation (1) & (2), the value of tensions \(T_1 \) & \(T_2\) will be obtained.

A \(T_1 = 13 \; N, \; T_2 = 3\; N\)

B \(T_1 = 10 \; N,\;\; T_2 = 5 \; N\)

C \(T_1 = 12 \; N, \;\;T_2 = 6 \; N\)

D \(T_1 = 5 \; N, \;\;\; T_2 = 10 \; N\)

- To avoid skidding of vehicle at the turn, the roads are banked at some angle.
- For this, the outer part of the road is lifted up, as compared to the inner part, as shown in figure.

\(\sum F_C = N \; sin\,\theta \)

Using Newton's second law,

\(N \; sin\, \theta = \dfrac{mv^2}{R}\) ...(1)

where R is the radius of curvature

v is the speed of vehicle

m is the mass of vehicle

\(N \; cos\, \theta -mg = 0\)

\(N \; cos\, \theta = mg \) ...(2)

- From equation (1) & (2)

\(tan\,\theta = \dfrac{v^2}{Rg}\)

\(\theta = tan^{-1} \left(\dfrac{v^2}{Rg}\right)\)

where \(\theta \) is the angle of banking of road.

- Thus, velocity of vehicle

\(v = \sqrt{Rg \; tan \,\theta}\)

- Consider a mass m, suspended from ceiling by a light string moving along a horizontal circle, as shown in figure. This arrangement is known as conical pendulum.
- The radius of the circle is R.
- Let the speed of the mass is v.

**\(\sum F_C = T sin\,\theta\)**

By Newton's second law,

\(Tsin\,\theta = ma_C \)

\(\Rightarrow \; T sin\, \theta = \dfrac{mv^2}{R} \) ...(1)

Acceleration = 0 in vertical direction

\(\Rightarrow \; T cos \; \theta - mg = 0\)

\(\Rightarrow \; T cos \; \theta = mg \) ...(2)

- From (1) & (2)

\(\Rightarrow \; tan\, \theta = \dfrac{v^2}{ Rg}\)

\(v = \sqrt {Rg\; tan\,\theta} \)

- From equation \( (2)\)

\(T \; cos\, \theta = mg\)

\(T = \dfrac{mg}{cos\, \theta }\)

- From FBD

\(tan\,\theta = \dfrac{R}{h},\;\;\;\;\;\;\;\;\;\;\;cos\, \theta = \dfrac{h}{\sqrt {R^2 + h^2}}\)

Thus, \(v= \sqrt {Rg \; tan\, \theta }\)

\(v=\sqrt{\dfrac{Rg× R}{h}}\)

Also, \(T = \dfrac{mg}{\dfrac{h}{\sqrt{R^2 + h^2 }}}\)

\(T = \dfrac{mg \sqrt {R^2 + h^2}}{h}\)

A \(10 \; N\)

B \(20 \sqrt 5\; N\)

C \(\sqrt 5\; N\)

D \(10 \sqrt 5\; N\)