Motion in Two Dimensions in Polar Coordinates | Mechanics & General Properties of Matter - Physics PDF Download

Application of Newton’s Law of Motion

To apply Newton’s law of motion one should follow following step:
Step 1 - Draw free body diagram and identified external forces
Step 2 - Write down equation of constraint.
Step 3 - Write down Newton’s law of motion.
Example 11: The force F in figure is just sufficient to hold the 100N block and weightless pulley in equilibrium. If There is no appreciable friction.

Then find T₁, T₂ and T₃
For plank T₁ +T₂ +T₃ = 100
F = T₁
For pulley 1 T₂ =T₁ + F = T₂ = T₁ + T₁ ⇒ T₂ =2T₁
For pulley 2 T₃ = 2T₂ = 4T₁
T₁  + 2T₁ + 4T₁ = 100 ⇒ T₁ = 100/7 = F = 100/7, T2 = 2T₁ ⇒ T₂ = 200/7 and T₃ = 4T₁/7 = 400/7

Example 12: The following parameters of the arrangement of are available: the angle α which the inclined plane forms with the horizontal, and the coefficient of friction k between the body m₁ and the inclined plane. The masses of the pulley and the threads, as well as the friction in the pulley, are negligible. Assuming both bodies to be motionless at the initial moment, find the mass ratio m₂/m₁ at which the body m₂
(a) starts coming down;(b) starts going up;(c) is at rest.

(a) for m₂ Starts coming down. For mass m₂, m₂g > T and for mass m₁ moving up T >m₁g sin α + f_max
therefore m₂g > m₁g sin α + f_max
m₂g > m₁g sin α + k₁m₁g cos α
m₂/m₁ > sin α+ k cos α
(b) For m₂ starts coming up. For mass m₂, m₂g < T and for mass m₁ moving up T <m₁g sin α - f_max
m₁g > m₁g sin α + f_max
m₂g > m₁g sin α + k₁m₁g cos α
m₂/m₁ > sin α+ k cos α

(c) At rest :friction will be static:

sin α - k cos α < m₂/m₁< sin α + k cos α

Example 13: A block of mass m is placed on another block of mass M lying on a smooth horizontal surface. The coefficient of static friction between m and M is µ s . What is the maximum force that can be applied to m so that the blocks remains at rest relative to each other?

Imagine the situation when F is at its maximum value so that m is about to start slipping relative to M. The mass m tries to drag M towards right due to friction.
Equation of motion of mass m:
F - µ_sN = ma
N = mg
Hence frictional force on M exerted by m will be towards right.
Let a = magnitude of acceleration of blocks towards right.
Equation of motion of mass  M:
µ_sN = MA ⇒ α = μ_smg/M R = N+ Mg
Solving these equations, we get:
F = µ_sN + ma ⇒ μ_smg + m.

If F is less than critical value, the blocks stick together without any relative motion.
If F is greater than this critical value, the blocks slide relative to each other and their accelerations are different.

Example 14: The blocks of masses m and M are not attached to each other but are in contact. The coefficient of static friction between the blocks is µ but the surface beneath M is smooth. What is the minimum magnitude of the horizontal force F required to hold m against M ?

If m and M sticking together they will have same acceleration.
Let a = acceleration of blocks
Equation of motion for mass m
F - R = ma and  f = mg f is frictional force
Equation of motion for mass M
R = Ma
f + Mg = N

Solve to get:
f = mg and R = MF/m + m

for no slipping,  f < µ_SR mg ≤ MF/M + m ⇒ F ≥

Example 15: At the moment t = 0 the force F = at is applied to a small body of mass m resting on smooth horizontal plane (a is a constant). The permanent direction of this force forms an angle α with the horizontal. Find
(a) the velocity of the body at the moment of its breaking off the plane;
(b) the distance traversed by the body up to this moment.

(a) If one will draw the free body diagram F sin α +N= mg in y direction F cos α = ma₁ in x direction, where a₁ is acceleration of block.
At time of breaking off the plane vertical component of must be equal to weight mg.

Then, F sin α = mg = at sin α sin ⇒ t = mg/a sin α
Equation of motion of block:
F cos α = m a₁ , a₁ =

(b) 

Example 16: In the arrangement shown in figure the bodies have masses m₀, m₁, m₂, the friction is absent, the masses of the pulleys and the threads are negligible. Find the acceleration of the body m_1.Look into possible cases.
 

T = 2T₁ and from equation of constrain a0 = a₁ + a₂/2
Equation of motion:


from (i), (ii) and (iii)

Example 17: Coefficient of friction is µ. What will be acceleration of table such that system will be in equilibrium?
 

This whole wedge is moving with acceleration a₀. Equation of motion for mass m which on the mass M
N = mg,
ma₀ + fr = T ⇒  ma₀ + µN₁ = T
ma₀ + µ mg = T ⇒ m(a₀ + μg) = T (1)

Equation of motion for mass m which is hanged vertical  the mass M 20

N₂ - mα0 = 0 ⇒ N₂ = mα₀
fr + T = mg ⇒ T = mg - fr, ⇒ T = mg - µmα0 ⇒ T = m(g - µα₀) (2)
From (1) and (2) ma₀ + mµg - mg + mµa₀ = 0
- g(1 - µ) + α₀ (µ + 1) = 0 ⇒ α₀ = g(1 - µ)/(1 + µ)

Example 18: In the arrangement shown in figure block of mass m slides on the surface of Wedge with inclination
. The masses of pulley and thread is negligible and friction on each surface is absent. If mass of Wedge is M then find the acceleration of wedge M.

Let us assume the mass M is moving with acceleration a₀ towards the wall . Hence the length of thread is fixed then the acceleration of the block m is also a₀ with respect to surface of wedge  in downward direction.

Now let’s draw free body diagram for mass m.
1. The weight mg in downward direction
2. The tension T along negative x axis
3. The normal force N along positive y axis.
4. If observer is attached to surface of wedge then he is on accelerating frame then he will measure Pseudo force ma0 in horizontal direction.
Now resolve all forces along suitable axis and write equation of motion
The equation of motion along x - axis is mg sin α + ma₀ cos α - T = ma₀
T = mg sin α + ma₀ (1 - cos α)…..(1)
The equation of motion along y axis
N + ma₀ sin α - mg cos α  = 0
⇒ N = mg cos α - ma₀ sin α ….(2)
Now draw free body diagram for wedge M
1. The weight Mg in downward direction
2. Normal force N₁ in y direction
3. Reaction force N due to block m on wedge
4. Hence pulley is attached on wedge so tension on thread pull the wedge along x axis and parallel to surface of wedge
Now resolve the forces along suitable axis
The equation of motion along x axis
T - T cos α + N sin α = Ma₀ ⇒ T =
Put the value of N from equation (2)
....(3)
The equation of motion along y - axis is N₁ - Mg + N cos α  + T sin α = 0
From equation (1) and equation (3)
⇒ T = mg sin α + mα₀ (1 - cos α) =
Now α₀ can be easily solve α₀ =

Motion in Two Dimensional in Polar Coordinate

Two dimensional motion in Cartesian coordinate

The position vector in two dimension ( x,y ) plane is given by  whereare unit vector in x and y direction respectively.
The base unit vector  are not vary with position as shown in figure.
The velocity is given by  and acceleration is given by Newton’s law can be written as
Two dimensional motion in polar coordinate Two dimensional system also can be represent in polar coordinate with variable (r,θ) with transformation rule x =r cosθ and y = r sin θ where r = where r identified as magnitude of vector and tan^-1 (y/x) and θ is angle measured from x axis in anti clock wise direction as shown in figure.
In polar coordinate system  are unit vector in radial direction and tangential direction of trajectory. One can see from the figure the  are vary with position, where conclude they are orthogonal in nature.
The unit vector  can be written in basis of unit vector .

The unit vectors  at a point in the xy -plane. We see that the orthogonality of andplus the fact that they are unit vectors,
which is shown.
The transformation can be shown by rotational Matrix


Time evolution ofand

One can easily see unit vectorandare vary with time.

The Position Vector in Polar Coordinate

r = ris sometimes confusing, because the equation as written seems to make no reference to the angle θ. We know that two parameters needed to specify a position in two dimensional space (in Cartesian coordinates they are x and y ), but the equation r = r seems to contain only the quantity r . The answer is thatis not a fixed vector and we need to know the value of θ to tell howis origin. Although θ does not occur explicitly in r, its value must be known to fix the direction of. This would be apparent if we wrote to emphasize the dependence ofon θ. However, by common conversation is understood to stand for(θ).

Velocity Vector in Polar Coordinate

Acceleration Vector in Polar Coordinate

is radial acceleration and a_θ = is tangential acceleration.
So Newton’s law in polar coordinate can be written as
F_r = ma_r = mwhere F_r is external  force in radial direction.

F_θ = ma_θ = where F_θ is external  force in tangential direction.

Example 19: A bead moves along the spoke of wheel at constant speed u meter per second .The wheel rotates with uniform angular velocity = ω radians per second about an axis fixed in space. At t = 0 the spoke is along the x axis and bead is at the origin.
(a) Find the velocity of particle
(b) Find the acceleration of particle

Circular Motion

For circular motion for radius r₀, at r = r₀, then  so F_r = ma_r = where F_r is force in radial direction and F_θ = ma_θ = , where F_θ is force in tangential direction.

There are two type of circular motion.
(1) Uniform Circular Motion
If there is not any force in tangential direction F_θ = 0 is condition, then motion is uniform circular motion i.e.,is constant known as angular speed and tangential speed is given by v =r₀ω.

(2) Non-uniform Circular Motion
For non-uniform circular motion of radius r radial acceleration is

tangential acceleration aθ dv/dt
the acceleration for non uniform circular motion  is given by
the magnitude of acceleration is given by

Example 20: Find the magnitude of the linear acceleration of a particle moving in a circle of radius 10 cm with uniform speed completing the circle in 4s.

The distance covered in completing the circle is 2πr = 2π × 10cm. The linear speed is

The linear acceleration is

This acceleration is directed towards the centre of the circle.

Example 21: A particle moves in a circle of radius 20 cm . Its linear speed is given by v = 2t , where t is in second and v in metre/second. Find the radial and tangential acceleration at t = 3s.

The linear speed at t = 3s is v = 2t = 6 m/s
The radial acceleration at t = 3s is
 
The tangential acceleration is

Calculation of torque and angular momentum

Torque is given by
Torque is also defined rate of change of momentum τ = DJ/dt = m
So, angular momentum is given by J =

If there is not any  tangential force is in plane then angular momentum of the system is conserve.

Example 22: The Spinning Terror
The Spinning Terror is an amusement park ride – a large vertical drum which spins so fast that everyone inside stays pinned against the wall when the floor drops away. What is the minimum steady angular velocity ω, which allows the floor to be dropped away safely?

Suppose that the radius of the drum is R and the mass of the body is M . Let µ be the coefficient of friction between the drum and M . The forces on M are the weight W , the friction force f and the normal force exerted by the wall, N as shown below. The radial acceleration is Rω² toward the axis, and the radial equation of motion is N =MRω²

By the law of static friction, f ≤ µN = µ MRω²
Since, we require M to be in vertical equilibrium, f = Mg,
and we have,  Mg ≤ µMRω² or ω² ≥ g/µR
The smallest value of ω that will work is

Example 23: A horizontal frictionless table has a small hole in the centre of table. Block A of mass m_a on the table is connected by a block B of mass m_b hanging beneath by a string of negligible mass can move under gravity only in vertical direction, which passes through the hole as shown in figure.
(a) Write down equation of motion in radial, tangential and vertical direction.
(b) What is equation of constrain?
(c) What will be the acceleration of B when A is moving with angular velocity ω at radius r₀ ?

Suppose of block A rotating in circle with angular velocity ω of radius r₀ what is acceleration of block B assumedirection is shown in figure.

The external force in radial direction is tension, not any external force in tangential direction and weight of body m_b and tension in vertical direction.
(a)


(b) Since length of the rope is constant then
 ....(4)

(c) Put value of

Example 24:  A bead rests at the top of a fixed frictionless hoop of radius R lies in vertical plane.
(a) At particular instance particle make angle θ with vertical Write down equation of motion in polar coordinate.
(b)  Find the value of θ = θ_c such that particle will just leave the surface of ring.
(c)  At what angle θ = θ_p the acceleration of the particle in vertical direction. What is relation between θ_c and θ_p.

(a) N - mg cosθ = ma_r =  and mg sinθ = ma_θ=
Equation of constrain

From conservation of energy



(c) a_r = 2g(1 - cosθ_P) and a_θ = g sinθ_P
If net acceleration is in vertical direction then horizontal component must be cancelled to each other.
a_r sinθ_P = a_θ cosθ_P ⇒2g(1 - cosθ_P)sinθ_P = g sinθ_P cosθ_P ⇒ cosθ_P = 2/3
θ_c = θ_P = cos^-1 2/3