Questions on Uniform Circular motion with answers

option b
Full form of rpm is revolution per minute. In circular motion the angular displacement for one complete revolution is 2\pi radian.
So, 1rpm = \frac{2\pi}{60} rad/sec

option d
In uniform circular motion, the speed (v) remains constant. So, the kinetic energy \frac{1}{2}mv^{2}, also remains constant. The magnitude of acceleration ( \frac{mv^{2}}{r}) also remains constant. But the direction of acceleration always changes. So, answer is option (d) all of these remain constant

option b
Frequency is defined as the number of cycles per unit time. Where as time period is the time for one complete cycle. If T is the time period, then in T sec 1 cycle is completed. So in 1 sec 1/T cycles will be completed and that is the frequency.
\nu = 1/T

option c
The relation between angular velocity and linear velocity is v = \omegar
The relation between angular acceleration ( \alpha) and linear acceleration (a) is a = \alphar ,
r is the radius of the circular path

option b
The minute hand completes one revolution in 60 min or 3600 sec. The angular displacement for one complete cycle is 2 \pi.
So, angular velocity of minute hand is
angular \hspace{0.2cm} velocity=\frac{angular \hspace{0.2cm} displacement}{time}
angular \hspace{0.2cm} velocity=\frac{2\pi}{3600} rad/s
angular \hspace{0.2cm} velocity=\frac{\pi}{1800}rad/s

When a particle moves in a uniform circular motion, a force acts towards the centre of he circular path. This force is called centripetal force. If m is the mass of the particle moving with uniform velocity v in a circular path of radius r, then the centripetal force is
F=m\frac{v^{2}}{r}

The relation between angular velocity ( \omega) and linear velocity v is v = \omegar, where r is the radius of the circular path.
Centripetal force,
F=m\frac{v^{2}}{r}
Put the value of v = \omegar
F=m\frac{(\omega r)^{2}}{r}
F=m\frac{\omega^{2} r^{2}}{r}
F=m\omega^{2} r

Centrifugal force is not a real force. It exists only in non-inertial reference frame or rotating frame of reference. It can not be observed from inertial frame of reference.

Banking of road is necessary to supply the required centripetal force to the vehicles in a turn. The outer edge of the road is raised above the inner edge of the road in a turn of a road. The angle at which the outer edge is raised is known as angle of banking.

In the figure, the angle of banking is \theta.
If the radius of the circular turn is r, m is the mass of the vehicle and R is the normal reaction of the road.
The weight of the vehicle mg acts downward.
The normal reaction R is resolved in two directions.
The vertical component RCos\theta balances the weight of the body.
The horizontal component RSin\theta provides the necessary centripetal force to take a safe turn.
Then RCos\theta = mg
RSin\theta =\frac{mv^{2}}{r}
m is the mass of the vehicle and v is the velocity.
\frac{RSin\theta}{RCos\theta}=\frac{\frac{mv^{2}}{r}}{mg}

tan\theta =\frac{v^{2}}{rg}
v=\sqrt{rgtan\theta }
here v is the maximum velocity at which a vehicle can take a safe turn.

A cyclist bends on a turning road to provide the centripetal force.
Let the cyclist bends at angle \theta with the vertical while taking a turn on a road.
The ground exerts a reactional force on the cyclist which will be an an angle \theta with the vertical.
Let m be the mass of the cycle with the cyclist

The vertical component RCos\theta balances the weight mg of the cyclist
The horizontal component RSin\theta gives the centripetal force to make the turn.
Then Rcos\theta = mg and
RSin\theta =\frac{mv^{2}}{r}
here v is the velocity of the cyclist
\frac{RSin\theta}{RCos\theta}=\frac{\frac{mv^{2}}{r}}{mg}

Angular displacement for 1 revolution is 2\pi rad
So, Angular displacement for 8 revolution is 2\pi \times 8=16\pi rad
Angular displacement = 16\pi rad
Time = 2 s
Angular velocity, \omega = \frac{Angular displacement}{time}
Angular velocity, \omega = \frac{16\pi}{2} = 8\pi rad/s
radius r = 1m
linear velocity, v = \omega r = 8\pi \times 1 = 8\pi m/s

mass of the object, m = 0.25 kg
radius, r = 200 mm = 0.2 m
Angular velocity, \omega = 1800 rpm
\omega=1800 \times \frac{2\pi}{60}=60\pi rad/s
Centripetal force, F=m\omega^{2}r
F=0.25 \times (60\pi)^{2} \times 0.2
F = 29.6 N

mass of the stone, m = 2 kg
length of the string, l = 1 m
Tension, T = 200 N
Let v is the maximum speed that the stone can have so that the string remains intact.
Here the tension supplies the required centripetal force to the stone for the whirling motion. Then
\frac{mv^{2}}{l}=T
v^{2}=\frac{lT}{m}
v=\sqrt{\frac{lT}{m}}
v=\sqrt{\frac{1\times 200}{2}}
v=10 m/s
So, the maximum speed of the stone can be 10 m/s

mass of the person, m = 60 kg
radius of curvature of the road, r = 10 m
velocity of the car, v = 50 m/s
Centrifugal force, F = \frac{mv^{2}}{r}
F = \frac{60 \times (50)^{2}}{10}
F=15000 N = 1.5 kN

radius of the curved railway line, r = 50 m
Speed of the train, v = 36 km/h = \frac{36\times 1000}{3600} m/s = 10 m/s
Let the angle of banking be \theta
Then, tan\theta =\frac{v^{2}}{rg}
tan\theta =\frac{(10)^{2}}{50 \times 9.8}
tan\theta =0.204
\theta = 11.53^{0}

radius of the curved road, r = 100 m
Speed of the train, v = 72 km/h = \frac{72\times 1000}{3600} m/s = 20 m/s
Let the angle of inclination to the vertical be \theta
Then, tan\theta =\frac{v^{2}}{rg}
tan\theta =\frac{(20)^{2}}{100 \times 9.8}
tan\theta =0.408
\theta = 22.195^{0}<