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Circular Motion Problems Without Friction

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.

Centripetal Acceleration

  • 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. 

Centripetal Force

  • 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. 

Net force along radial direction

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

By Newton's second law,

\(Tsin\,\theta = ma_C\)

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

Net force along vertical direction

Acceleration = 0 in vertical direction 

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

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

  • From (1) & (2)

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

\(v = \sqrt {Rg\; tan\,\theta} \)  

Illustration Questions

A ball of mass \(m = 2\; kg\) suspended from ceiling by a light string, moving along a horizontal circle of radius \(R = 1\; m\), as shown in figure. If string traces a cone of height \(h = 0. 5 \; m\), find the speed of ball.  Given : \(g = 10\; m/s\)

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

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

C \(5 \; m/s\)

D \(10 \; m/s\)

×

Net force in radial direction:

\(\sum F_C = Tsin\, \theta\)

Using Newton's second law,

\(Tsin\, \theta = ma_C\)

[where aC is centripetal acceleration]

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

[let v be the speed of the ball]

Net force in vertical direction:

Acceleration = 0 in vertical direction

\(Tcos\,\theta - mg = 0\)

\(Tcos\, \theta = mg\)     ...(2)

From equation (1) & (2)

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

image

From trigonometry,

\(tan\,\theta = \dfrac{\text{perpendicular}}{\text{Base}}\)

\(tan\,\theta = \dfrac{1}{0.5}\)

\(tan\,\theta = 2\)      

\(2 = \dfrac{v^2}{Rg}\)

\(v = \sqrt{2\; Rg}\)

\(= \sqrt {2 × 1 × 10}\)

\(= \sqrt {20}\)       

\(= 2 \sqrt 5 \; m/s\)      

A ball of mass \(m = 2\; kg\) suspended from ceiling by a light string, moving along a horizontal circle of radius \(R = 1\; m\), as shown in figure. If string traces a cone of height \(h = 0. 5 \; m\), find the speed of ball.  Given : \(g = 10\; m/s\)

image
A

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

.

B

\(\sqrt10\; m/s\)

C

\(5 \; m/s\)

D

\(10 \; m/s\)

Option A is Correct

Angle of Banking of Roads

  • 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.

Net radial force towards center

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

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

Net force in vertical direction

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

\(N \, cos \,\theta = mg\)     ...(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. 

Illustration Questions

The road has a circular turn of radius \(R = 1.6 \; m\). The vehicle is moving with velocity \(v = 4 \; m/s\). What should be the angle of banking? Given : \(g = 10\; m/s^2\)

A 15°

B 30°

C 20°

D 45°

×

The angle of banking of road is given by

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

 

Given : \(R = 1.6 \; m,\;\;\;\;v = 4\; m/s,\;\;\;\;\;g = 10 \; m/s^2\)

\(\theta = tan ^{-1} \; \left(\dfrac{(4)^2}{(1.6)× (10)}\right)\)

\(\theta =tan ^{-1} \;\;\;(1)\)

\(\theta = 45°\)

The road has a circular turn of radius \(R = 1.6 \; m\). The vehicle is moving with velocity \(v = 4 \; m/s\). What should be the angle of banking? Given : \(g = 10\; m/s^2\)

A

15°

.

B

30°

C

20°

D

45°

Option D is Correct

Calculation of Speed for Banked Roads

  • 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.

Net radial force towards center

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

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 

Net force in vertical direction

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

\(N \; cos\, \theta = mg \)   ...(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 

\(v = \sqrt{Rg \; tan \,\theta}\)    

Illustration Questions

A road at a circular turn of radius \(R = 1.6 \; m\) is banked by \(\theta = 45° \). What should be the speed of vehicle at the turn so that normal contact force is able to provide the necessary centripetal force?    Given : \(g = 10\; m/s^2\) 

A 2 m/s

B 3 m/s

C 4 m/s

D 5 m/s

×

Let v be the speed of vehicle. The angle of banking of road is given by 

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

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

Given : \(R = 1.6 \; m,\;\;\;\;g = 10\; m/s^2,\;\;\;\;\;\theta = 45 ° \)

\(v = \sqrt{1.6 × 10 × tan \; 45° }\)

\(v = \sqrt {16}\)

\(v = 4 \; m/s\)

A road at a circular turn of radius \(R = 1.6 \; m\) is banked by \(\theta = 45° \). What should be the speed of vehicle at the turn so that normal contact force is able to provide the necessary centripetal force?    Given : \(g = 10\; m/s^2\) 

A

2 m/s

.

B

3 m/s

C

4 m/s

D

5 m/s

Option C is Correct

Calculation of Different Parameters for Given Conical Pendulum 

  • 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. 

Net force along radial direction

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

By Newton's second law,

\(Tsin\,\theta = ma_C \)

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

Net force along vertical direction

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

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

Illustration Questions

A ball of mass \(m = 2 \; kg\) suspended from ceiling by a light string, moving along a horizontal circle of radius \(R = 1 \; m\), as shown in figure. If string traces a cone of height \(h = 0. 5\; m\), find the tension in the string.  Given : \(g = 10 \; m/s^2\)

A \(10 \; N\)

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

C \(\sqrt 5\; N\)

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

×

Net force along radial direction

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

By Newton's second law,

\(Tsin\,\theta = ma_C \)

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

Net force along vertical direction

Acceleration = 0 in vertical direction 

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

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

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

\(T = \dfrac{mg \sqrt {R^2 +h^2}}{h}\)                  \(\left[\because\;cos\, \theta \;\dfrac{h}{\sqrt{R^2 +h^2}}\right]\)       

image

Given: \(m = 2 \; kg , \;\;\;\;\;\;g = 10 \,m/s^2, \;\;\;\;\;\;\;\;\;R = 1\, m,\;\;\;\;\;h = 0.5 \,m\)

\(T = \dfrac{2 × 10 \sqrt {(1)^2 + (0.5)^2}}{0.5}\)

\(= 20 \sqrt 5\; N\)      

A ball of mass \(m = 2 \; kg\) suspended from ceiling by a light string, moving along a horizontal circle of radius \(R = 1 \; m\), as shown in figure. If string traces a cone of height \(h = 0. 5\; m\), find the tension in the string.  Given : \(g = 10 \; m/s^2\)

image
A

\(10 \; N\)

.

B

\(20 \sqrt 5\; N\)

C

\(\sqrt 5\; N\)

D

\(10 \sqrt 5\; N\)

Option B is Correct

Calculation of Minimum Speed to Cross the Top Most Point of Path of String in Vertical Circular Motion  

  • 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. 

Illustration Questions

A ball is attached with a light string and moving in a vertical circular motion. If ball is moving in a circle of radius \(R = 0. 2\; m\), then find the minimum speed of the ball to cross the top most point without slacking of string.  Given :\(\,\,\,\,g = 10\,m/s^2\)

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

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

C \(2 \; m/s\)

D \(3 \; m/s\)

×

The minimum speed of the ball to cross the top most point without slacking of string is 

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

Given : \(R = 0.2\,m , \;\;\;\;\;\;\;\;g = 10 \; m/s^2\)

\(v_{min} = \sqrt {0.2 × 10 }\)

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

A ball is attached with a light string and moving in a vertical circular motion. If ball is moving in a circle of radius \(R = 0. 2\; m\), then find the minimum speed of the ball to cross the top most point without slacking of string.  Given :\(\,\,\,\,g = 10\,m/s^2\)

A

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

.

B

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

C

\(2 \; m/s\)

D

\(3 \; m/s\)

Option A is Correct

Analysis of Circular Motion When a Ball is Attached with Two Strings 

  • 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.

Net force along radial direction

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

Using Newton's second law,

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

Net force in vertical direction 

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

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

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

Illustration Questions

A ball of mass \(m = \dfrac{1}{\sqrt 2}\; kg \) attached with two strings moving in circular path of radius of radius \(R = 1 \; m\) with velocity \(v = 4 \; m/s\), as shown in figure. Find the tensions in the strings. \(\theta _1 = 45 ° , \;\;\;\theta _2 = 45 °\) Given: \(g = 10 \; m/s^2\) 

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

×

Net force along radial direction

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

Using Newton's second law,

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

Net force in vertical direction 

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

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

image

Given: \(m = \dfrac{1}{\sqrt 2}\; kg, \;\;\;\;\;\;\;\;\;R = 1 \; m,\;\;\;\;\;\;\;\;\;v=4 \; m/s \)

\(g=10 \; m/s^2, \;\;\;\;\;\;\theta_1=\theta _2 = 45° \)

From equation (1),

  \(T_1 \; cos\; 45° + T_2 \; cos \; 45° = \dfrac{\left(\dfrac{1}{\sqrt 2}\right)(4)^2}{1}\)

\(T_1 \left(\dfrac{1}{\sqrt 2}\right) + T_2 \left(\dfrac{1}{\sqrt 2}\right) = \left(\dfrac{1}{\sqrt 2}\right) (16)\)

\(T_1 + T_2 = 16\)      ...(3)

From equation (2),

\(T_1 \; sin \; 45° - T_2 \; sin \; 45° = \left(\dfrac{1}{\sqrt 2}\right) 10\)

\(T_1 \left(\dfrac{1}{\sqrt2}\right) - T_2\left(\dfrac{1}{\sqrt2}\right) = \dfrac{10}{\sqrt 2} \)

\(T_1 - T_2 = 10\)       ...(4)

Adding (3) & (4)

\(T_1 + T_2 +T_1 -T_2 = 16 +10\)

\(2T_1 = 26 \)

\(T_1 = 13 \; N\)

Putting value of \(T_1\) in \((3)\)

\(13 +T_2 = 16\)

\(T_2 = 3 \; N\)        

A ball of mass \(m = \dfrac{1}{\sqrt 2}\; kg \) attached with two strings moving in circular path of radius of radius \(R = 1 \; m\) with velocity \(v = 4 \; m/s\), as shown in figure. Find the tensions in the strings. \(\theta _1 = 45 ° , \;\;\;\theta _2 = 45 °\) Given: \(g = 10 \; m/s^2\) 

image
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

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

Option A is Correct

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