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\$

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>>Class 11
>>Physics
>>Laws of Motion
>>Dynamics of Circular Motion
>> $Circular Turning and Banking of Roads$

Circular Turning and Banking of Roads

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 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 :

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\$

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 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

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)\$

Q: 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 ?

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\$

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\$

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: 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 :

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

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Important Questions

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)

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

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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 :

Medium

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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

Easy

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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

Easy

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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 ?

Medium

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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

Hard

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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.

Medium

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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.

Medium

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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 :

Medium

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Circular Turning and Banking of Roads

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Circular Motion and Centripetal Force

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Circular Motion

Circular Motion and Centripetal Force

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)

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

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 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 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 ?

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

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 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.

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

A 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 :