Q: One end of string of length \$l\$ is connected to a particle of mass '\$m\$' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed '\$v\$', the net force on the particle (directed towards center) will be (\$T\$ represents the tension in the string)

\$ N=mg\$ & \$T=\dfrac{mv^2}{r}\$Hence the resultant force is T. inwords

Q: A \$500\ kg\$ car takes a round turn of radius \$50\ m\$ with a velocity of \$36\ km/hr\$. The centripetal force is

Given : \$m = 500 kg\$ \$R = 50\$ mVelocity of the car \$ v = 36 km/hr = 36 \times \dfrac{5}{18} = 10 m/s\$ \$\therefore\$ Centripetal force \$F = \dfrac{mv^2}{R} = \dfrac{500 \times 10^2}{50} = 1000 N\$

Q: A string of length, \$L\$, is fixed at one end and carries a mass, \$M\$ at the other end. The string makes \$2/\pi\$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

Given : \$w = \dfrac{2}{\pi}\$ rev. per sec \$\implies w = 2\pi \nu = 4\$From figure \$R = L sin\theta\$\$\therefore\$ \$T sin\theta = MR w^2\$\$\implies\$ \$T = M L w^2 = ML (4)^2 =16 ML\$

Q: Suppose the gravitational force varies inversely as the \$n\$th power of distance, then the time period of a planet in circular orbit of radius \$R\$ around the sun will be proportional to.

F=\$\dfrac{GmM}{R^n}=m\omega^2R\rightarrow\omega=\sqrt{\dfrac{GM}{R^{n+1}}}\$ hence time period is proportional to \$R^{\dfrac{n+1}{2}}\$

Q: An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Radius of the loop, \$r = 1 km = 1000 m\$Speed of the aircraft,\$ v = 900 km/h = 900 \times 5 / 18 = 250 m/s\$Centripetal acceleration, \$a_c = \dfrac {v^2 }{ r} \$ \$= \dfrac{(250)^2}{ 1000} = 62.5 m/s^2\$Acceleration due to gravity, \$g = 9.8 m/s^2\$Ratio: \$\dfrac{a_c}{ g} = 62.5 / 9.8 = 6.38\$ \$a_c = 6.38 g\$

Question 1

What is banking of roads? Obtain an expression for the maximum safety speed of a vehicle moving along a curved horizontal road.

Accepted Answer

Banking of roads : To avoid risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inwards, i.e., the outer side of road is raised above its inner side. This is called 'banking of roads'.Consider a car taking a left turn along a road of radius $r$ banked at an angle $\theta$ for a designed optimum speed $V$. Let $m$ be the mass of the car. In general, the forces acting on the car are:(a) Its weight $\vec {mg}$, acting vertically down(b) The normal reaction of the road $\vec {N}$, perpendicular to the road surface(c) The frictional force $\vec {f_{s}}$ along the inclined surface of the road.Resolve $\vec {N}$ and $\vec {f_{s}}$ into two perpendicular components $N\cos \theta$ vertically up and $\vec {f_{s}} \sin \theta$ vertically down, $N\sin \theta$ and $\vec {f_{s}} \cos \theta$ horizontally towards the centre of the circular path.If $v_{max}$ is the maximum safe speed without skidding.$\dfrac {mv_{max}^{2}}{r} = N\sin \theta + f_{s}\cos \theta$               $= N\sin \theta + \mu_{s} N\cos \theta$$\dfrac {mv_{max}^{2}}{r}= N(\sin \theta + \mu_{s} \cos \theta) .... (1)$and $N\cos \theta = mg + f_{s} \sin \theta$               $= mg + \mu_{s} N\sin \theta$   $\therefore mg = N(\cos \theta - \mu_{s}\sin \theta) ... (2)$Dividing eq. $(1)$ by eq. $(2)$,$\dfrac {mv_{max}^{2}}{r.mg} = \dfrac {N(\sin \theta + \mu_{s}\cos \theta)}{N(\cos \theta - \mu_{s}\sin \theta)}$$\therefore \dfrac {v_{max}^{2}}{rg} = \dfrac {\sin \theta + \mu_{s}\cos \theta}{\cos \theta - \mu_{s}\sin \theta} = \dfrac {\tan \theta + \mu_{s}}{1 - \mu_{s}\tan \theta}$$\therefore v_{max} = \sqrt {\dfrac {rg(\tan \theta + \mu_{s})}{1 - \mu_{s}\tan \theta}} ...,. (3)$This is the expression for the maximum safe speed on a banked road.At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting $\mu_{s} = 0$ in eq. $(3)$, the optimum speed on a banked circular road is$v = \sqrt {rg \tan \theta} ... (4)$$\therefore \tan \theta = \dfrac {v^{2}}{rg}$ or $\theta = \tan^{-1}\left (\dfrac {v^{2}}{rg}\right )$From this eq. we see that $\theta$ depends upon $v, r$ and $g$. The angle of banking is independent of the mass of a vehicle negotiating the curve.

Question 2

One end of string of length $l$ is connected to a particle of mass '$m$' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed '$v$', the net force on the particle (directed towards center) will be ($T$ represents the tension in the string)

Accepted Answer

$ N=mg$ & $T=\dfrac{mv^2}{r}$Hence the resultant force is T. inwords

Question 3

A $500\ kg$ car takes a round turn of radius $50\ m$ with a velocity of $36\ km/hr$. The centripetal force is

Accepted Answer

Given :      $m = 500  kg$                         $R =  50$ mVelocity of the car        $ v  = 36  km/hr  =  36 \times \dfrac{5}{18}  = 10  m/s$                  $\therefore$ Centripetal force      $F  = \dfrac{mv^2}{R}  = \dfrac{500 \times 10^2}{50}  = 1000  N$

Question 4

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place on a plane. It follows that

Accepted Answer

Since the particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, so work done by the force in the direction of the velocity is zero. So the magnitude of velocity will not change due to the force. Kinetic energy is $KE=\dfrac{1}{2}mv^{2}=constant $.Now mass and magnitude of velocity will always be constant in this given situation so the kinetic energy will also be constant.Hence, the correct option is $(C)$

Question 5

A coin placed on a rotating turntable just slips if it is placed at a distance of $4$ cm from centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of :

Accepted Answer

When the coin is just on the verge of slipping, it just overcomes the force of static friction (${ F }_{ s }$) . which acts as centripetal force for being that coin in the circular motion.So we have ${ F }_{ s }=m{ \omega  }^{ 2 }r$ If we double the $\omega$,  then $r$ will also self adjust to a new ${ r }^{ ' }$, since the static friction will remain same. We have $\quad m{ \omega  }^{ 2 }r=m{ (2\omega ) }^{ 2 }{ r }^{ ' }\ \Rightarrow { r }^{ ' }=\dfrac { r }{ 4 } \ cm=\dfrac { 4 }{ 4 }\  cm=1\ cm$

Question 6

A string of length, $L$, is fixed at one end and carries a mass, $M$ at the other end. The string makes $2/\pi$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

Accepted Answer

Given :        $w = \dfrac{2}{\pi}$ rev. per sec        $\implies w = 2\pi \nu  = 4$From figure       $R  = L sin\theta$$\therefore$    $T sin\theta  = MR w^2$$\implies$       $T   = M L w^2  = ML (4)^2   =16 ML$

Question 7

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Accepted Answer

Net acceleration is due to braking and centripetal accelerationDue to Braking,$a_T= 0.5 m/s^2$Speed of the cyclist, $v = 27 km/h = 7.5 m/s$Radius of the circular turn, $r=80m$Centripetal acceleration is given as:$a_c= \dfrac{V^2}{r}$  $=(7.5 ^2)/80= 0.70 m/s^2$Since the angle between $a_c$ and $a_T$ is 900, the resultant acceleration a is given by:$a= (a_c^2+a_T^2)^{1/2} $$a=(0.7^2+ 0.5^2)^{1/2}= 0.86 m/s^2$$tan \theta= \dfrac{a_c}{a_T}$where $\theta $ is the angle of the resultant with the direction of the velocity. $tan \theta=\dfrac{0.7}{0.5}= 1.4 $$\theta = tan^{-1}(1.4) = 54.56^0$

Question 8

Suppose the gravitational force varies inversely as the $n$th power of distance, then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to.

Accepted Answer

F=$\dfrac{GmM}{R^n}=m\omega^2R\rightarrow\omega=\sqrt{\dfrac{GM}{R^{n+1}}}$ hence time period is proportional to $R^{\dfrac{n+1}{2}}$

Question 9

An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Accepted Answer

Radius of the loop, $r = 1 km = 1000 m$Speed of the aircraft,$ v = 900 km/h = 900 \times 5 / 18  =  250 m/s$Centripetal acceleration, $a_c = \dfrac {v^2 }{ r} $                                               $= \dfrac{(250)^2}{ 1000} = 62.5 m/s^2$Acceleration due to gravity, $g = 9.8 m/s^2$Ratio:             $\dfrac{a_c}{ g} = 62.5 / 9.8  =  6.38$                        $a_c = 6.38 g$

Question 10

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of $10\ ms^{-1}$. It implies that the boy is :

Accepted Answer

As an object moves in a merry-go-round, the speed of the object may be constant, but the direction of the object changes with the motion of the object hence its velocity will also change.As velocity is changing, therefore the boy will be in an accelerated motion.Hence option C is correct

Circular Turning and Banking of Roads

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REVISE WITH CONCEPTS

Circular Motion and Centripetal Force

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Circular Motion

Circular Motion and Centripetal Force

What is banking of roads? Obtain an expression for the maximum safety speed of a vehicle moving along a curved horizontal road.

A $500 k g$ car takes a round turn of radius $50 m$ with a velocity of $36 k m / h r$ . The centripetal force is

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place on a plane. It follows that

A coin placed on a rotating turntable just slips if it is placed at a distance of $4$ cm from centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of :

A string of length, $L$ , is fixed at one end and carries a mass, $M$ at the other end. The string makes $2 / π$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Suppose the gravitational force varies inversely as the $n$ th power of distance, then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to.

An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of $10 m s^{- 1}$ . It implies that the boy is :

Circular Turning and Banking of Roads

Language

Rate

REVISE WITH CONCEPTS

Circular Motion and Centripetal Force

QUICK SUMMARY WITH STORIES

Circular Motion

Circular Motion and Centripetal Force

What is banking of roads? Obtain an expression for the maximum safety speed of a vehicle moving along a curved horizontal road.

One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v', the net force on the particle (directed towards center) will be (T represents the tension in the string)

A 500 kg car takes a round turn of radius 50 m with a velocity of 36 km/hr. The centripetal force is

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place on a plane. It follows that

A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of :

A string of length, L, is fixed at one end and carries a mass, M at the other end. The string makes 2/π revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Suppose the gravitational force varies inversely as the nth power of distance, then the time period of a planet in circular orbit of radius R around the sun will be proportional to.

An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms−1. It implies that the boy is :

A $500 k g$ car takes a round turn of radius $50 m$ with a velocity of $36 k m / h r$ . The centripetal force is

A coin placed on a rotating turntable just slips if it is placed at a distance of $4$ cm from centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of :

A string of length, $L$ , is fixed at one end and carries a mass, $M$ at the other end. The string makes $2 / π$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

Suppose the gravitational force varies inversely as the $n$ th power of distance, then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to.

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of $10 m s^{- 1}$ . It implies that the boy is :