Notes
Topics:
- Angular Displacement Angular Velocity and Angular Acceleration
- Relation between Linear and Angular Velocities
- Tangential and Radial Accelerations
- Uniform Circular Motion
- Centripetal and Centrifugal Forces
- Banking of a Curved Road
- Vertical Circular Motion due to Earth's Gravitation
- Kinematic Equations for Circular Motion in Analogy with Linear Motion
Motion : Motion is the change in position of an object with time. When a particle moves in a circular path, then its motion is said to be circular motion.
Circular motion is two dimensional motion. A particle moving in a circular path shows many of the important features of the velocity and acceleration vectors in two dimensions.
When the speed of the particle performing circular motion is constant, then its motion is said to be uniform circular motion.
If the speed of the particle performing circular motion is changing, then its motion is said to be non-uniform circular motion.
Angular Displacement : It is defined as the angle turned by the particle from some reference line. Angular displacement is usually measured in radians. It is dimensionless quantity.
Finite angular displacement is a scalar but an infinitesimally displacement is a vector. Since, it does not obey the commutative law of vector addition.
Angular Velocity : It is defined as the rate of change of the angular displacement of the body undergoing circular motion. A body executing circular motion Angular velocity ω = \(\lim\limits_{Δt\to0}\frac{Δθ}{Δt}=\frac{dθ}{dt}\) It is an axial vector whose direction is given by the right hand rule. Its unit is rad s-1 and dimension is [T-1]. Note If a body complete one revolution in time T, then its angular velocity. ω = \(\frac{2π}{t}\) It a body complete n revolution in one second, then its angular velocity. ω =2πn Instantaneous Angular Velocity : It is defined as the limiting value of average angular velocity of the particle in a small time interval, as the time interval approaches zero. ωinst = \(\lim\limits_{Δt\to0}\frac{Δθ}{Δt}=\frac{dθ}{dt}\) Instantaneous angular velocity ωinst= \(\frac{dθ}{dt}\) The angular velocity is perpendicular to the plane of circular motion, it is directed upwards for anticlockwise circular motion and directed downwards for clockwise circular motion. Linear Velocity : It is defined as the rate of change linear displacement \(v=\frac{Δs}{Δt}\) if Δt → 0 \(v=\lim\limits_{Δt\to0}\frac{Δs}{Δt}=\frac{ds}{dt}\) It is a vector quantity and its unit is ms-1 and its dimension is [LT-1]. Relation between Linear and Angular Velocities : When a particle is moving along a curved path, then its tangential and angular velocities are related by \(v=ωr\) The direction of v, and r are mutually perpendicular. Where vector r is joining the location of particle and the point about which w has been computed.
Angular Acceleration : It is defined as the rate of change of angular Velocity of a body. Let ω1, and ω2 be the instantaneous angular speed s. At times t1 and t2 respectively, then the average angular acceleration is defined as
\(a_ω=\frac{ω_1-ω_2}{t_2-t_1}=\frac{Δω}{Δt}\)
It is a vector quantity and its unit is rad s-1 and dimension is [T-1]
Instantaneous Angular Acceleration : The angular acceleration of a body at a given instant of time or at a given point of its motion is called its instantaneous angular acceleration.
Thus, \(α_{inst}=\lim\limits_{Δω\to0}\frac{Δs}{Δt}=\frac{dω}{dt}\)
\(α_{inst}=\frac{d^2θ}{dt^2} (....where ω=\frac{dθ}{dt})\)
Acceleration : Acceleration of a particle in circular motion has two following components.
Two kinds of acceleration of a body executing circular motion
i. Tangential Acceleration (at): It is the component of acceleration a in the direction of velocity.
at = component of a along v = dv/dt = d|v|/dt
ii Radial Acceleration (ar): It is the component of a towards the centre of the circular motion. This is responsible for a change in the direction of velocity.
ar = v2/r = rω2 = 4π2n2r
Radial acceleration is also called centripetal acceleration.
As at and ar are perpendicular to each other as shown in the figure.
Finding the resultant of the acceleration
∴ Net acceleration, a = \(\sqrt{a_t^2+a_r^2}\) ; tanθ = \(\frac{a_t}{a_r}\)
If a particle is moving on a circular path with a constant speed, then
at = dv/dt = 0 but ar ≠ 0
Uniform Circular Motion : If a particle in circular motion moves with an uniform speed, then motion of the particle is called uniform circular motion. In such a case, v is constant, then
dv/dt=0 and a = ω2 r = \(\frac{v^2}{r^2}⋅r=\frac{v^2}{r}\)
Thus, if a particle moves in a circle of radius r with an uniform speed v, then its acceleration is v2/r This acceleration is termed as centripetal acceleration and it is always directed towards the centre of the circular path.
Force in Circular Motion : When a body executes circular motion, then there are two types of forces as given below:
Centripetal Force : The centripetal force is the force required to move a body along a circular path with a constant speed. Centripetal force never acts by itself. It is to be provided by some agency in order to maintain the uniform circular motion of an object. The direction of the centripetal force is along the radius, acting towards the centre of the circle, on which the given body is moving.
Centripetal force, F = \(\frac{mv^2}{r}\) = mrω2
F = mr4π2v2
Where, m= mass of body
v= speed of the body
r = radius of circural path
ω = angular speed
[It is a vector quantity. Its unit is Newton and dimension is [MLT-2].
Centripetal Force in Different Situations
| Situation | The Centripetal Force . |
| A particle tied to a string and whirled in a horizontal circle | Tension in the string |
| Vehicle taking a turn on a level road | Frictional force exerted by the road on the tyres |
| Revolution of the earth around the sun | Gravitational force exerted by the sun |
| Electron revolving around the nucleus in an atom | Coulomb attraction exerted by the protons on electrons |
| A charged particle describing a circular path in a magnetic field | Magnetic force exerted by the magnetic field |
| A vehicle on a speed breaker | Weight of the body or a component of weight |
Centrifugal Force : Centrifugal force is a virtual force due to incorporated effects of inertia, The centrifugal force is a pseudo force experienced by a non-inertial observer moving in a circular path with a constant speed on account of its directional inertia, Mathematically,
Centripetal force = \(\frac{mv^2}{r}\) = mrω2
Skidding of Vehicle on a Level Road :
When a vehicle takes a turn on a circular path it requires centripetal force.
If friction provides this centripetal force then vehicle can move in circular path safely i.e.
Friction force ≥ Required centripetal force
μmg ≥ \(\frac{mv^2}{r}\) ⇒ vsaf = \(\sqrt{μrg}\)
Banking of a Curved Road :
Friction is not always reliable at circular turns, if high speed and sharp turns are involved.
To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat raised compared to the inner part.
Let the road be banked at an angle q from the horizontal, as shown in the figure.
If b = width of the road and h = height of the outer edge of the road as compared to the inner edge, then
tanθ =\(\frac{v^2}{rg}=\frac hb\)
In such a case, when friction is present between road and tyre. Then,
Maxim speed, \(v_{max}=\sqrt\frac{rg(μ_s+tanθ)}{1-μ_stanθ}\)
Where μs = coefficient of static friction
Conical Pendulum : Conical pendulum is a simple pendulum, which is given with such a motion that bob describes a horizontal circle and the string describes a cone. It consists of a string OA, whose upper end O is fixed and bob is tied at the other free end A.
Let the bob is moving in a horizontal circle of radius r with an uniform angular velocity in such a way that the string always makes an angle q with the vertical. As the string traces the surface of the cone, the arrangement is called a conical pendulum.
Forces on a conical pendulum executing circular motion
Let T be the tension in the string of length \(l \) and r the radius of circular path.
The vertical component of tension T balances the weight of the bob and horizontal component provides the necessary centripetal force.
Thus, Tcosθ = mg ……………(i)
and Tsinθ = mrω2 …………..(ii)
On dividing Eq. (ii) by Eq. (i), we get
tanθ = \(\frac{rω^2}{g}\)
i.e. ω =\(\sqrt\frac{gtanθ}{r}\)
but r= l sinθ and ω=2π/T
where, T being the period of completing one revolution,
∴ 2π/T \(\sqrt\frac{gtanθ}{l sinθ}\)
T= 2π \(\sqrt\frac{lsinθ}{g\frac{sinθ}{cosθ}}\)
Or Period T = 2π \(\sqrt\frac{lcosθ}{g}\)=\(2π\sqrt\frac{h}{g}\)
Where h= l cosθ
Vertical Circular Motion due to Earth's Gravitation : In vertical plane, acceleration due to gravity (g) acts on a particle in circular motion. Hence, the motion is not uniform. As work is done against the gravitational field the particle must be given minimum energy, so that it rotates in circular motion.
Equation for Velocity and Energy at Different Positions of a Vertical Circular Motion
It is an example of non-uniform circular motion in which speed of object decreases due to effect of gravity as the object goes from its lowest position A to highest position B.
Force on an object executing circular motion at two different points
The following are the observations taken from the object executing vertical circular motion :
i, At the lowest point A, the tension TL and the weight mg are in mutually opposite directions and their resultant provides the necessary centripetal force,
i.e. TL – mg = \(\frac{mv_L^2}{r}\)
Tension at lowest point TL = mg + \(\frac{mv_L^2}{r}\)
ii, At the highest point B, tension TH and the weight mg are in the same direction and hence,
TH + mg =\(\frac{mv_H^2}{r}\)
Tension at highest point TH = \(\frac{mv_H^2}{r}\) – mg
Moreover, vL and vH are correlated as \(v_H^2= v_L^2-4gr\)
iii In general, if the revolving particle, at any instant of time, is at position C, inclined at an angle q from the vertical, then
\(v^2=v_L^2-2gr(1-cosθ)\)
Tension at any instant, T = mg cosq +\(\frac{mv^2}{r}r\)
iv. In the critical condition of just looping the vertical loop, (i.e. when the tension just becomes zero at the highest point B, we obtain the following results
TH = 0, TL = 6mg,
\(v_L=\sqrt{5rg}\) and \(v_H=\sqrt{rg}\)
In general, TL — TH =6mg.
Kinematic Equations for Circular Motion in Analogy with Linear Motion :
There are three kinematic equation in linear motion,
V=u+at, s=ut+ (½)at2 and v2 =u2 +2as
where, v= final velocity, u = initial velocity,
s = distance traveled, t = time taken
a = linear acceleration
Consider a body rotating with initial angular velocity (ωo) and having constant angular acceleration (α). The kinematical equations for its circular motion are given by
ω = ωo + αt ...(i)
θ= ωot + ½ αt2 ...(ii)
ω2 = ω2o + 2 αθ
where, w= final angular velocity,
ωo = initial angular velocity
θ = angular displacement
α = angular acceleration
t = time taken
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