Informative line

### 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)

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

$$v = \sqrt {Rg\; tan\,\theta}$$  #### 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 $$10 \; m/s$$

×

$$\sum F_C = Tsin\, \theta$$

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)

$$tan \,\theta = \dfrac{v^2}{Rg}$$ 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 : $$g = 10\; m/s$$ 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.  #### 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 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.

#### 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.  #### 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) 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)

$$T_1 + T_2 +T_1 -T_2 = 16 +10$$

$$2T_1 = 26$$

$$T_1 = 13 \; N$$

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: $$g = 10 \; m/s^2$$ 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

# 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

$$v = \sqrt{Rg \; tan \,\theta}$$  #### 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,         $$T = \dfrac{mg}{\dfrac{h}{\sqrt{R^2 + h^2 }}}$$

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

$$T = \dfrac{mg}{cos\, \theta }$$

$$T = \dfrac{mg \sqrt {R^2 +h^2}}{h}$$                  $$\left[\because\;cos\, \theta \;\dfrac{h}{\sqrt{R^2 +h^2}}\right]$$ 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 : $$g = 10 \; m/s^2$$ A

$$10 \; N$$

.

B

$$20 \sqrt 5\; N$$

C

$$\sqrt 5\; N$$

D

$$10 \sqrt 5\; N$$

Option B is Correct