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Circular Motion
Circular motion:
Motion of a body in circular orbit is called circular motion. For example; motion of planets and satellites in their orbits, motion of electrons around the nucleus of an atom, motion of different hands of watch etc.
Terms used in circular motion:
Angular displacement:
The angle swept out by the radius vector of the particle at the centre of circular orbit in given interval of time is called angular displacement. Generally it is denoted by θ and its SI unit is radian. It is a vector quantity. Its direction is along the axis of rotation perpendicular to the plane of motion, given by right hand rule. So, it is an axial vector.
If a body moves from point A to B in time interval t, then 𝜃 represents the angular displacement of time interval t.
Angular velocity:
The rate of change of angular displacement with time is called angular velocity. It is denoted by 𝜔. Its SI unit is radian per second (rad/sec). It is also an axial vector directed along the axis of rotation according to the right-hand rule.
If 𝜃 be the angular displacement of a body in time interval t, then average angular velocity is given by
𝝎 = 𝜽𝒕
Alternatively, if θ 1 and θ 2 be the angular displacements of a body in time interval t 1 and t 2 from initial position, the average angular velocity between the time t 1 to t 2 is given by
𝜔 =
𝜃 2 −𝜃 1 𝑡 2 −𝑡 1 =^
Δ𝜃 Δ𝑡 (where,^ Δ𝜃=^ 𝜃^2 − 𝜃^1 &^ Δ𝑡^ =^ 𝑡^2 − 𝑡^1 )
The instantaneous angular velocity 𝜔 is given by the limiting value of average acceleration.
i.e., ω =
Δ𝜃 Δ𝑡
⇒ 𝛚 =
𝐝𝜽 𝐝𝒕
B
O θ^ A
Δ𝑡 → 0
lim
i.e., the derivative of angular displacement with respect to time gives the instantaneous angular velocity.
For uniform circular motion, the average and instantaneous values of angular velocity are same.
Angular acceleration:
The rate of change of angular velocity is called angular acceleration. It is also an axial vector having a unit rad/s^2. It is denoted by 𝛼.
If the angular velocity of a body changes from initial value 𝜔𝑜 to final value 𝜔 in time interval t, then average angular acceleration is given by
𝜶 = 𝝎 − 𝝎 𝒕 𝒐
Alternatively, if 𝜔 1 and 𝜔 2 are the angular velocities of a body in time t 1 and t 2 , then average angular velocity between the time t 1 to t 2 is given by
𝛼 = 𝜔 𝑡^2 − 𝜔^1 2 −𝑡 1
The instantaneous angular acceleration is given by the limiting value of average angular velocity.
i.e., 𝛼 = Δ𝜔Δ𝑡
⇒ 𝜶 = 𝐝𝝎𝐝𝒕
i.e., the derivative of angular velocity with respect to time gives the instantaneous angular acceleration.
Time period:
Time period of a particle moving in a circular path is defined as the time taken by the particle to complete one revolution. It is denoted by T. Its SI unit is second.
∵ 𝝎 = 𝜽𝒕 and for t = T, 𝜃 = 2𝜋,
𝜔 = 2𝜋𝑇
∴
Frequency:
The frequency of a particle moving in a circular path is defined as the number of revolutions made by the particle in one second. It is denoted by f.
It is given by the relation,
f =
𝟏 𝑻 (Its units are revolution/sec or cycle/sec or Hertz) ∵ 𝜔 = 2𝜋𝑇 = 2𝜋 × (^) 𝑇^1
∴
Δ𝑡 → 0
lim
T = 𝟐𝝅 𝝎
Expression for centripetal acceleration:
Consider a particle moving in a circular orbit in the XY plane with centre at O and radius r with uniform speed v. Let P (x,y) be the position of the particle at any time t, and the corresponding angular displacement is 𝜃. The angular velocity of the particle is given by
∴ 𝜃 = 𝜔𝑡 ……………………(i) The position vector of the particle at time t can be expressed as, 𝑟⃑ = x 𝑖̂ + y 𝑗̂ or, 𝑟⃑ = rcos𝜃 𝑖̂ + r𝑠𝑖𝑛𝜃 𝑗̂ = r (cos𝜔𝑡 𝑖̂ + 𝑠𝑖𝑛𝜔𝑡 𝑗̂ ) ……………(ii) Now, the velocity of the particle is given by,
𝑣⃑ = 𝑑𝑟⃑𝑑𝑡 = r𝜔 (-sin𝜔𝑡 𝑖̂ + cos𝜔𝑡𝑗̂ )
Also the acceleration of the particle is given by
𝑎⃑ = 𝑑𝑣⃗⃑𝑑𝑡 = 𝑑[ r𝜔 (−sin𝜔𝑡 𝑖̂ + 𝑐𝑜𝑠𝜔𝑡 𝑗̂ )]𝑑𝑡 = - r𝜔^2 (cos𝜔𝑡 𝑖̂ + 𝑠𝑖𝑛𝜔𝑡 𝑗̂)
or, 𝑎⃑ = - 𝜔^2 r (cos𝜔𝑡 𝑖̂ + 𝑠𝑖𝑛𝜔𝑡 𝑗̂ ) or, 𝑎⃑ = - 𝜔^2 𝑟⃑ ………….(iii) This is the expression for centripetal acceleration. Negative sign shows that the direction of centripetal acceleration is opposite to the position vector 𝑟⃑. That is towards the centre of the circle. In magnitude, a = 𝜔^2 r
Also in terms of linear velocity, a = 𝑣
2 𝑟 (∵ v = ωr)
𝑌′
𝑣⃑
P(x,y)
Y
𝑋′ O X
𝜃 x
y 𝑟⃑
Alternative method for centripetal acceleration:
Consider a body moving in a circular orbit having centre at O and radius r with uniform speed v. Let v⃗⃗⃗A⃗ and v⃗⃗⃗B⃗ are the velocities of the body at points A and B and angle subtended by the arc AB at the centre of the circle is Δθ as shown in fig i. To find the change in velocity in going from A to B, the velocities - v⃗⃗⃗⃗A and v⃗⃗⃗⃗B are represented by vectors 𝑃𝑄⃗⃗⃗⃗⃗ and 𝑅𝑃⃗⃗⃗⃗⃗ as shown in fig ii. The
resultant of the vectors 𝑅𝑃⃗⃗⃗⃗⃗ and 𝑃𝑄⃗⃗⃗⃗⃗ is represented by the vector 𝑅𝑄⃗⃗⃗⃗⃗ which represents the
change in velocity v⃗⃗⃗B⃗ - v⃗⃗⃗⃗A = ∆v⃗⃗⃗⃗. Let ∆𝑡 be the time taken by the body in going from A to B. For very small interval of time , i.e., for ∆𝑡 → 0, the points A and B are very close to each other and the angle Δθ is also small. For Δθ → 0, the change velocity ∆v⃗⃗⃗⃗ is perpendicular to both v⃗⃗⃗⃗A and v⃗⃗⃗B⃗ and is towards the centre of the circle.
The magnitude of acceleration is given by
a = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = Δ𝑣Δ𝑡
For small value of Δ𝜃,
Δθ = RQRP = Δvv [ ∵ | 𝑅𝑃⃗⃗⃗⃗⃗|= |𝑃𝑄⃗⃗⃗⃗⃗|= v ]
i.e., Δv = v Δθ
∴ a = v Δ𝜃Δ𝑡
For Δt → 0,
a = Δt → 0 v Δ𝜃Δ𝑡 = v dθdt [ ∵ v is constant ]
∴ a = v𝜔 = 𝑣 × 𝑣𝑟
Fig : ii
𝑣⃗⃗⃗𝐴⃗
𝑣⃗⃗⃗𝐵⃗ B
O A
Δθ
r
Fig : i
a = 𝑣
2 𝑟
R
Q
P
𝑣⃗⃗⃗𝐵⃗ 𝑣⃗⃗⃗𝐴⃗
Δθ
Applications of centrifuge: a. Cream is separated from milk in cream separator. b. Wet clothes are dried in drying machine. c. Centrifuges are used in sugar industry to separate sugar crystal from molasses. d. Centrifuges are used to separate honey from wax. Applications of centripetal forces
1. Motion of a vehicle in a level curved path:
Consider a vehicle of mass m moves in a level curved path of radius r with speed v. The centripetal force between the tyre and the road acts towards the centre of the circle and provides necessary centripetal force.
i.e., Fc = Fr
or, 𝑚𝑣
2 𝑟 =^ 𝜇𝑅^ ( where^ 𝜇^ is the coefficient friction between the tyre and the road ) or, 𝑚𝑣
2 𝑟 =^ 𝜇^ mg ∴ v = (^) √𝝁 𝐫𝐠
This is the maximum velocity with which the vehicle can take a turn in a level curved path of radius r.
mg
R
mg
𝐹𝑟
Fig : Motion of vehicle in a level curved path
2. Banking of road :
The phenomenon of raising of outer edge of the road over the inner edge in a curved path is called banking of road. The angle by which outer edge is raised over the inner edge is called banking angle or angle of banking. Roads are banked in curved path to make the centripetal force independent to the force of friction.
Consider a vehicle moving with speed v in a curved path of radius r having banking angle 𝜃. In banked road, the normal reaction R of the ground to the vehicle makes same angle 𝜃 with the vertical. The normal reaction R is resolved into two components; the vertical component Rcos𝜃 balances the weight mg of the vehicle and horizontal component Rsin𝜃 provides necessary centripetal force. i.e., Rcos𝜃 = mg ......................(i)
And, Rsin𝜃 = 𝑚𝑣
2 𝑟 ....................(ii) Dividing eqn^ (ii) by (i) ,we get
tanθ = v
2 rg
or, θ = tan- |^ (v
2 rg ) ..............................(iii) This is the expression for the angle by which the road should be banked in a curved path of radius r to move safely with speed v. Also the safe limit of velocity can be expressed as, v = (^) √rgtanθ ...........................(iv)
mg
R
mg
Rcosθ
Rsinθ
Fig : Banking of road
4. Motion in a horizontal circle - Conical Pendulum:
Consider a body of mass m is tied to a string of length l and whirled round a horizontal circle of radius r with uniform speed v. Let θ be the angle made by the string with vertical. When the mass moves in a horizontal circle, the motion of the string describes a cone. So it is also called conical pendulum. When the mass moves in a horizontal circle, the tension ‘T’ in the string acting towards the point of suspension can be resolved into two components. The vertical component Tcosθ balances the weight mg of the mass and the horizontal component Tsinθ provides necessary centripetal force. i.e., Tcos𝜃 = mg ......................(i)
And, Tsin𝜃 = 𝑚𝑣
2 𝑟 ....................(ii) Dividing eqn^ (ii) by (i) ,we get
tan𝜃 = 𝑣
2 𝑟𝑔 or, v = √rgtanθ
or, ω r = (^) √rgtanθ ( where ω is angular velocity of the mass)
or, ω = √gtanθr
Now the time period of conical pendulum is given by
t = 2πω
or, t = 2π √g tanθr
= 2π√ (^) g tanθr
From figure; sinθ = rl ⇒ r = lsinθ
mg
Tcosθ T
Tsinθ
Fig : Conical Pendulum
∴ t = 2π√ (^) g tanθl sinθ = 2π√l sinθ ×cosθg sinθ
or,
This is the expression for the time period of conical pendulum.
- Motion in a Vertical circle:
Consider a body of mass m is tied to one end of a string and whirled round a vertical circle of radius r with speed v. When the body moves in the vertical circle, the tension T in the string always acts towards the centre of the circle. Let at any instant, the mass is at point P and the angle made by the string with vertical is 𝜃. When the mass is at point P, the component of weight mg of the mass acting opposite to the tension is mg cos𝜃. So, the resultant force acting towards the centre of the circle is T – mg cos𝜃, which provides the necessary centripetal force for the mass to move in the circular path.
i.e., T – mg cos𝜃 = 𝑚𝑣
2 𝑟 or, T = 𝑚𝑣
2 𝑟 + mg cos𝜃^ ......................(i)
This is the expression for the tension in the string; when a mass tied to a string is rotated in a vertical circle.
Cases:
i. When the mass is at lowest point of vertical circle (at point A), 𝜃 = 0𝑜
∴ T = 𝑚𝑣
2 𝑟 + mg
t = 2π√l^ cos gθ
Fig : Motion in a vertical circle
D C
B
A
P 𝜃
mg^ mg cos𝜃
T
Similarly, the minimum velocity of mass when the string becomes horizontal is given by
v 3 = (^) √𝟑𝐫𝐠 ........................(iii) Conceptual questions:
- Is it necessary that the acceleration in circular motion is always towards the centre of the circle? Explain. Ans : There are two types of acceleration in circular motion. One is centripetal acceleration or radial acceleration (𝑎⃗⃗⃗⃗𝑐 ), which acts towards the centre of the circle and keeps the body in circular path. Another is tangential acceleration (𝑎⃗⃗⃗𝑇⃗ ) which changes the speed of the body. So, the acceleration is circular motion is given by the resultant of these two accelerations. i. e., 𝑎 = 𝑎⃗⃗⃗⃗𝑐 + 𝑎⃗⃗⃗𝑇⃗ Cases : i. If 𝑎⃗⃗⃗⃗𝑇 = 0 , 𝑎⃗⃗⃗⃗𝑐 ≠ 0, then 𝑎 = 𝑎⃗⃗⃗⃗𝑐 i.e., the acceleration is towards to centre of the circle and the motion is uniform circular motion. ii. If 𝑎⃗⃗⃗⃗𝑇 ≠ 0 , 𝑎⃗⃗⃗⃗𝑐 ≠ 0, then 𝑎 = 𝑎⃗⃗⃗⃗𝑐 + 𝑎⃗⃗⃗⃗𝑇 with magnitude ; 𝑎 = √𝑎𝑐^2 + 𝑎𝑇^2 i.e., the acceleration is not towards to centre of the circle and the motion is non uniform circular motion. iii. If 𝑎⃗⃗⃗⃗𝑇 ≠ 0 , 𝑎⃗⃗⃗⃗𝑐 = 0, then 𝑎 = 𝑎⃗⃗⃗⃗𝑇 and the motion is linear but not circular Thus, in case of uniform circular motion, the acceleration is towards the centre of the circle and in case of non-uniform circular motion it is not towards the centre of the circle.
- Does centripetal force do a work? Explain. Ans : Work done by a force is given by W = FS cosθ In case of centripetal force, the force and displacement are perpendicular. i.e. θ = 90 𝑜 ∴ W = FS cos 90o^ = 0 Thus, the centripetal force does not do any work.
- Why is it more difficult to revolve a stone by tying it to a longer string than a shorter string? Ans : To move a body in circular path , a force towards the centre called centripetal force is required which is given by, F = m𝜔^2 r. Thus to revolve a stone in longer string, a larger centripetal force is required. So, it is more difficult to revolve a stone by tying it to a longer string than a shorter string.
- The positively charged nucleus of an atom attracts the electrons in the orbit. Why do the electrons not collapse into the nucleus?
𝑎⃗⃗⃗𝑇⃗
𝑎 𝑎⃗⃗⃗⃗𝑐
Ans: The electrostatic force between the nucleus and electrons is used to provide necessary centripetal force for the electrons to move in circular orbit. In case of centripetal force, the force and displacement are perpendicular to each other and the force cannot do any work to pull the electrons towards the centre (W = FS cos90o^ = 0). So, the electrons do not collapse into the nucleus.
- Why does a cyclist incline himself with vertical while taking turn in a curved path?
Ans : When he inclines himself with vertical, the normal reaction of the ground makes some angle with vertical. The vertical component of normal reaction balances the weight and horizontal component provides necessary centripetal force to move in the curved path as shown in fig. Numerical Probems:
- A bob of mass 200 gm is whirled in a horizontal circle of radius 50 cm by a string inclined at 30o^ to the vertical. Calculate the tension in the string and the speed of the bob in the horizontal circle. (Ans: 2.3 N, 1.7 m/s)
- An object of mass 0.5 kg is rotated in a horizontal circle by a string 1m in length. The maximum tension in the string before it breaks is 50 N. What is the greatest number of revolution per second of the object? (Ans: 1.6 rev/s)
- A mass of 0.2 kg is rotated by a string at constant speed in a vertical circle of radius 1 m. If the minimum tension in the string is 3 N, calculate the magnitude of speed and the maximum tension in the string. (Ans: 5 m/s, 7 N)
- At what angle should a circular road be banked so that a car running at 50 km/hr is safe to go round the circular turn of 200 m radius? ( Ans: 5.5o^ )
- A certain string breaks when a weight of 25 N acts on it. A mass of 500 gm is attached to one end of the string of 1 m long and is rotated in a horizontal circle. Find the greatest number of revolutions per minute which can be made without breaking the string. (Ans: 67.5 rpm)
- A bob of mass 1 kg is attached to the lower end of the string 1 m long whose upper end is fixed. The mass is made to rotate in a horizontal circle of radius 60 cm. If the circular speed of the mass is constant, find the tension in string and the period of motion. (Ans: 12.5 N, 1.77 s)
- A coin is placed on a disc rotating with speed of 33 13 rev/min provided that the coin is not more than 10 cm from the axis. Calculate the coefficient of static friction between the coin and the disc. ( Ans: 0.12 )
mg
Rsinθ
Rcosθ R
a) 2π rad/sec b) 2π rad/min c) 2π rad/day d) 𝝅𝟔 rad/hr
- A car is moving with speed 30 m/s on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m/s^2. What is the acceleration of the car? a) 2 m/s^2 b) 2.7 m/s^2 c) 1.8 m/s^2 d) 9.8 m/s^2
- What is the ratio of angular speed of minute hand and second hand of a clock? a) 1 : 12 b) 12 : 1 c) 1 : 60 d) 60 : 1
- A particle is moving along a circular path. The angular velocity, linear velocity, angular acceleration and centripetal acceleration are denoted by 𝜔⃗⃗ , 𝑣 , 𝛼 and 𝑎⃗⃗⃗⃗𝑐 , which of the following statement is not correct? a) 𝜔⃗⃗ ⊥ 𝑎⃗⃗⃗⃗𝑐 b) 𝝎⃗⃗⃗ ⊥ 𝜶⃗⃗ c) 𝜔⃗⃗ ⊥ 𝑣 d) 𝑣 ⊥ 𝑎⃗⃗⃗⃗𝑐
- When a body moves with a constant speed along a circle: a) Its velocity remains constant b) No force acts on it c) No work is done on it d) No acceleration is produced in it
- A string can withstand a tension of 25 N, What is the greatest speed at which a body of mass 1kg can be whirled in a horizontal circle using 1m length of the string? a) 2.5 m/s b) 5 m/s c) 7.5 m/s d) 10m/s
- A body crosses the topmost point of a vertical circle with critical speed. What will be its acceleration when the string is horizontal? a) g b) 2g c) 3g d) 6g
- An object is tied to a string and rotated in a vertical circle of radius r. Constant speed v is maintained along the trajectory. If Tmax/ Tmin = 2, then v^2 /rg is : a) 1 b) 2 c) 3 d) 4
- Figure shows a body of mass m moving with a uniform speed v along a circle of radius r. The change in speed in going from C to D is :
a) v b) v/2 c) v/√2 d) Zero
- In question 11, change in velocity in going from A to B is :
a) v √𝟐 b) v c) v/√2 d) Zero
B
D
C (^) A
