Informative line

### Circular Motion Problems Involving Friction

Practice FBD for circular motion with friction, find maximum safe speed along horizontal circular rough road and on banked road for given coefficient of friction.

# FBD for Circular Motion with Friction

• Consider a car is moving on a circular path of radius $$R$$, as shown in figure.
• To provide enough centripetal force so that the car does not skid, the frictional force acts towards the center of radius of curvature.

• $$f_s$$ is called static friction and is self adjustable force. Its value is always less than or equal to $$\mu_sN$$, i.e., $$f_s \le \mu_sN$$.

#### A block is placed on a rough surface of a rotating table. The block is moving in circular motion. What will be the FBD of the block?

A

B

C

D

×

FBD of the block placed on a rotating table

To provide enough centripetal force such that the block does not slip, the frictional force acts towards the center of circular motion. This is called static friction and is self adjustable force. Its value is always less than or equal to $$\mu_sN$$ i.e., $$f_s \le \mu_sN$$.

Hence, option (A) is correct.

### A block is placed on a rough surface of a rotating table. The block is moving in circular motion. What will be the FBD of the block?

A
B
C
D

Option A is Correct

# Role of Friction for Safe Turn on Banked Road

• To avoid skidding of vehicle at the turn, the roads are banked at some angle.
• Even then, friction plays very important role for taking a safe turn.

• At the turn, the tyres get a tendency to skid outward along the lifted part.
• So, the frictional force which acts to oppose this skidding, works towards the center, as shown in figure.

• Applying Newton's second law along radial direction

$$Nsin\,\theta+fcos\,\theta=\dfrac{mv^2}{R}$$

where v is the velocity of vehicle

R is the radius of curvature

• Applying Newton's second law along vertical direction

$$Ncos\,\theta-fsin\,\theta-mg=0$$

• Dividing (1) by (2)

$$\dfrac{sin\,\theta+\mu_s\;cos\,\theta}{cos\,\theta - \mu_s\;sin\,\theta} =\dfrac{v^2}{Rg}$$  ... (3)

$$\Rightarrow sin\,\theta+\mu_s\; cos\,\theta = \dfrac{v^2}{Rg}cos\,\theta-\mu_s \dfrac{v^2}{Rg}sin\,\theta$$

$$\Rightarrow \mu_s = \dfrac{\dfrac{v^2}{Rg}cos\,\theta -sin\,\theta}{cos\,\theta+ \dfrac{v^2}{Rg}sin\,\theta}$$

$$\Rightarrow \mu_s= \dfrac{v^2cos\,\theta-Rg\;sin\,\theta}{v^2 sin\,\theta+Rg\;cos\,\theta}$$

#### A vehicle is moving on a circular banked track with the speed $$v=10\, m/s$$. If the angle of banking of road is $$\theta=37°$$ and radius of curvature of turn is $$R=10\ m$$, then find the coefficient of friction for the safe turn.  $$\Big[Given :g=10\ m/s^2,\,\,sin37°=\dfrac{3}{5},\,cos37°=\dfrac{4}{5}\Big]$$

A 0.25

B 0.14

C 0.36

D 0.48

×

At the turn, the tyres get a tendency to skid outward along the lifted part.

So, the frictional force which acts to oppose this skidding, works towards the center, as shown in figure.

Applying Newton's second law along radial direction

$$Nsin\,\theta+fcos\,\theta=\dfrac{mv^2}{R}$$

where v is the velocity of vehicle

R is the radius of curvature

Applying Newton's second law along vertical direction

$$Ncos\,\theta-fsin\,\theta-mg=0$$

Dividing (1) by (2)

$$\dfrac{sin\,\theta+\mu_s\;cos\,\theta}{cos\,\theta - \mu_s\;sin\,\theta} =\dfrac{v^2}{Rg}$$  ... (3)

$$\Rightarrow sin\,\theta+\mu_s\; cos\,\theta = \dfrac{v^2}{Rg}cos\,\theta-\mu_s \dfrac{v^2}{Rg}sin\,\theta$$

$$\Rightarrow \mu_s = \dfrac{\dfrac{v^2}{Rg}cos\,\theta -sin\,\theta}{cos\,\theta+ \dfrac{v^2}{Rg}sin\,\theta}$$

$$\Rightarrow \mu_s= \dfrac{v^2cos\,\theta-Rg\;sin\,\theta}{v^2 sin\,\theta+Rg\;cos\,\theta}$$

Given : $$v=10\ m/s\ ,\ \theta=37°\ ,\ R=10\ m$$

$$\mu_s=\dfrac{(10)^2cos\;37° - (10)(10)sin\;37°}{(10)^2sin\;37° + (10)(10)cos\;37°}$$

$$\mu_s=\dfrac{100\left[\dfrac{4}{5} - \dfrac{3}{5}\right]}{100\left[\dfrac{3}{5} + \dfrac{4}{5}\right]}$$

$$\mu_s=\dfrac{1}{7}$$

$$\mu_s=0.14$$

### A vehicle is moving on a circular banked track with the speed $$v=10\, m/s$$. If the angle of banking of road is $$\theta=37°$$ and radius of curvature of turn is $$R=10\ m$$, then find the coefficient of friction for the safe turn.  $$\Big[Given :g=10\ m/s^2,\,\,sin37°=\dfrac{3}{5},\,cos37°=\dfrac{4}{5}\Big]$$

A

0.25

.

B

0.14

C

0.36

D

0.48

Option B is Correct

# Maximum Safe Speed on Banked Road for Given Coefficient of Friction

• To avoid skidding of vehicle at the turn, the roads and banked at some angle.
• Even then, friction plays very important role for taking a safe turn.

• At the turn, the tyres get a tendency to skid outward along the lifted part.
• So, the frictional force which acts to oppose this skidding, works towards the center, as shown in figure.

• Applying Newton's second law along radial direction

$$Nsin\,\theta+fcos\,\theta=\dfrac{mv^2}{R}$$

where v is the velocity of vehicle

R is the radius of curvature

• Applying Newton's second law along vertical direction

$$Ncos\,\theta-fsin\,\theta-mg=0$$

• If the speed of vehicle is such that the limiting friction is applied then,

$$f_s=\mu_sN$$

$$Nsin\,\theta+\mu_sN cos\,\theta = \dfrac{mv^2}{R}$$ ... (1)

$$Ncos\,\theta-\mu_sN sin\,\theta = mg$$   ... (2)

Dividing (1) by (2)

$$\dfrac{sin\,\theta+\mu_s\;cos\,\theta}{cos\,\theta - \mu_s\;sin\,\theta} =\dfrac{v^2}{Rg}$$  ... (3)

$$v^2=\dfrac{Rg(sin\,\theta+\mu_s cos\,\theta)}{(cos\,\theta-\mu_s sin\,\theta)}$$

$$v_{max}=\sqrt{\dfrac{Rg(sin\,\theta+\mu_s cos\,\theta)}{(cos\, \theta-\mu_s sin\,\theta)}}$$

#### A vehicle is moving on a circular banked track whose banking angle is $$\theta=45°$$ and radius of curvature is $$R=\left(\dfrac{24}{5}\right)\ m$$. If the coefficient of friction is $$\mu_s = 0.5$$, then find the maximum safe speed of the vehicle at the turn.      $$\Big[ Given :g=10\ m/s^2\Big]$$

A 12 m/s

B 14 m/s

C 16 m/s

D 10 m/s

×

At the turn, the tyres get a tendency to skid outward along the lifted part.

So, the frictional force which acts to oppose this skidding, works towards the center, as shown in figure.

Applying Newton's second law along radial direction

$$Nsin\,\theta+fcos\,\theta=\dfrac{mv^2}{R}$$

where v is the velocity of vehicle

R is the radius of curvature

Applying Newton's second law along vertical direction

$$Ncos\,\theta-fsin\,\theta-mg=0$$

If the speed of vehicle is such that the limiting friction is applied then,

$$f_s=\mu_sN$$

$$Nsin\,\theta+\mu_sN cos\,\theta = \dfrac{mv^2}{R}$$ ... (1)

$$Ncos\,\theta-\mu_sN sin\,\theta = mg$$   ... (2)

Dividing (1) by (2)

$$\dfrac{sin\,\theta+\mu_s\;cos\,\theta}{cos\,\theta - \mu_s\;sin\,\theta} =\dfrac{v^2}{Rg}$$  ... (3)

$$v^2=\dfrac{Rg(sin\,\theta+\mu_s cos\,\theta)}{(cos\,\theta-\mu_s sin\,\theta)}$$

$$v_{max}=\sqrt{\dfrac{Rg(sin\,\theta+\mu_s cos\,\theta)}{(cos\, \theta-\mu_s sin\,\theta)}}$$

Given :  $$\theta = 45°,\ R=\dfrac{24}{5}\ m ,\ \mu_s=0.5$$

$$v_{max}=\sqrt{\dfrac{{\dfrac{24}{5}×10\bigg[sin45°+(0.5)cos45°\bigg]}}{cos45°-(0.5)sin45°}}$$

$$v_{max}=\sqrt{\dfrac{48\left[\dfrac{1}{\sqrt{2}}+(0.5)\dfrac{1}{\sqrt{2}}\right]}{\dfrac{1}{\sqrt{2}}-(0.5)\dfrac{1}{\sqrt{2}}}}$$

$$v_{max}=\sqrt{\dfrac{48×\dfrac{1}{\sqrt{2}}[1+0.5]}{\dfrac{1}{\sqrt{2}}[1-0.5]}}$$

$$v_{max}=\sqrt{\dfrac{72}{0.5}}=\sqrt{144}=12\ m/s$$

### A vehicle is moving on a circular banked track whose banking angle is $$\theta=45°$$ and radius of curvature is $$R=\left(\dfrac{24}{5}\right)\ m$$. If the coefficient of friction is $$\mu_s = 0.5$$, then find the maximum safe speed of the vehicle at the turn.      $$\Big[ Given :g=10\ m/s^2\Big]$$

A

12 m/s

.

B

14 m/s

C

16 m/s

D

10 m/s

Option A is Correct

# Maximum Safe Speed along Horizontal Circular Rough Road

• Consider a car is moving on a circular path of radius R, as shown in figure.
• To provide enough centripetal force so that the car does not skid, the frictional force acts towards the center of radius of curvature.

• This is called static friction and is self adjustable force. Its value is always less than or equal to $$\mu_sN$$, i.e., $$f_s \le \mu_sN$$.
• Thus, for a safe turn,

$$f_s=\dfrac{mv^2}{R}$$$$N=mg$$

• If $$\mu_s$$ is the coefficient of static friction between tyres and the road, then

$$f_s\le \mu_smg$$

• For safe turn,

$$\dfrac{mv^2}{R} \le \mu_s mg$$

$$\mu_s \ge\dfrac{v^2}{Rg}$$

$$v^2 \le \mu_s Rg$$

Maximum safe speed

$$v = \sqrt{\mu_s Rg}$$

#### A car of mass $$m=1000\ kg$$ is moving on a horizontal circular rough road with some velocity. If coefficient of static friction is $$\mu_s =0.5$$ and radius of curvature is $$R=20\ m$$, then find the maximum speed of the car for taking a safe turn.  $$\Big [ Given :g=10\ m/s^2\Big]$$

A $$10\ m/s$$

B $$15\ m/s$$

C $$20\ m/s$$

D $$30\ m/s$$

×

Maximum speed of the car at which it can take safe turn is given by

$$v=\sqrt{\mu_sRg}$$

Given : $$\mu_s =0.5, \ \ \ R=20\ m,\ \ \ g=10\ m/s^2$$

$$v=\sqrt{0.5×20×10}$$

$$=\sqrt{100}$$

$$=10\ m/s$$

### A car of mass $$m=1000\ kg$$ is moving on a horizontal circular rough road with some velocity. If coefficient of static friction is $$\mu_s =0.5$$ and radius of curvature is $$R=20\ m$$, then find the maximum speed of the car for taking a safe turn.  $$\Big [ Given :g=10\ m/s^2\Big]$$

A

$$10\ m/s$$

.

B

$$15\ m/s$$

C

$$20\ m/s$$

D

$$30\ m/s$$

Option A is Correct

# Friction for Static State

• Consider a rotor, a hollow vertical cylindrical structure which rotates about its axis.
• A person rests against the inner wall.
• At a particular speed of the rotor, the floor below the person falls off and the person hangs resting against the wall without any floor.
• Applying Newton's second law along radial direction

$$N=\dfrac{mv^2}{R}$$

• Newton's second law along vertical direction

$$f_s=mg$$

#### A block of mass $$m=2\ kg$$ is in static condition inside a moving hollow cylinder. If velocity of cylinder is $$v=5\ m/s$$, then find the friction force applied on the block.   $$\Big[Given : g=10\, m/s^2\Big]$$

A 30 N

B 10 N

C 20 N

D 40 N

×

FBD of block

Frictional force is applied on the block.

Applying Newton's second law along vertical direction

$$f_s=mg$$

Given : $$m=2\ kg,\ \ \ g=10\ m/s^2$$

$$f_s=2×10$$

= 20 N

### A block of mass $$m=2\ kg$$ is in static condition inside a moving hollow cylinder. If velocity of cylinder is $$v=5\ m/s$$, then find the friction force applied on the block.   $$\Big[Given : g=10\, m/s^2\Big]$$

A

30 N

.

B

10 N

C

20 N

D

40 N

Option C is Correct

# Angular Speed for Static State

• Consider a rotor, a hollow vertical cylindrical structure which rotates about its axis.
• A person rests against the inner wall.
• At a particular speed of the rotor, the floor below the person falls off and the person hangs resting against the wall without any floor.
• Applying Newton's second law along radial direction

$$N=\dfrac{mv^2}{R}$$

• Newton's second law along vertical direction

$$f_s=mg$$

• $$f_s$$ represents the static friction.
• Maximum value of static friction is given by,

$$f_s=\mu_sN$$

where $$\mu_s$$ is the coefficient of static friction.

• Thus, minimum speed for the block,

Putting value of $$f_s$$ and N in $$f_s=\mu_sN$$

$$mg=\mu_s \dfrac{mv^2}{R}$$

$$v=\sqrt{\dfrac{Rg}{\mu_s}}$$

#### In a rotor, a hollow vertical cylindrical structure rotates about its axis, a block is at rest against inner wall, as shown in figure. If the radius of the rotor is $$R=2\ m$$ and the coefficient of static friction is $$\mu_s=0.2$$, then find the minimum speed at which the block remains at rest even when the floor is removed. $$\Big[Given:g=10\ m/s^2\Big]$$

A 0.1 m/s

B 12 m/s

C 1 m/s

D 10 m/s

×

The minimum speed at which the block remains at rest even when the floor is removed, is given by

$$v=\sqrt{\dfrac{Rg}{\mu_s}}$$

Given : $$R=2\ m\ ,\ g=10\ m/s^2 ,\ \mu_s=0.2$$

$$v=\sqrt{\dfrac{2×10}{0.2}}$$

$$v=\sqrt{100}$$

= 10 m/s

### In a rotor, a hollow vertical cylindrical structure rotates about its axis, a block is at rest against inner wall, as shown in figure. If the radius of the rotor is $$R=2\ m$$ and the coefficient of static friction is $$\mu_s=0.2$$, then find the minimum speed at which the block remains at rest even when the floor is removed. $$\Big[Given:g=10\ m/s^2\Big]$$

A

0.1 m/s

.

B

12 m/s

C

1 m/s

D

10 m/s

Option D is Correct