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Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Hints on problem solving

1: Calculation of acceleration as time rate of change of speed gives tangential acceleration.

2: Calculation of acceleration as time rate of change of velocity gives total acceleration.

3: Tangential acceleration is component of total acceleration along the direction of velocity. Centripetal acceleration is component of acceleration along the radial direction.

4: We exchange between linear and angular quantities by using radius of circle, "r", as multiplication factor. It is helpful to think that linear quantities are bigger than angular quantities. As such, we need to multiply angular quantity by “r” to get corresponding linear quantities and divide a linear quantity by “r” to get corresponding angular quantity.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to non-uniform circular motion. The questions are categorized in terms of the characterizing features of the subject matter :

  • Velocity
  • Average total acceleration
  • Total acceleration

Velocity

Problem : A particle, tied to a string, starts moving along a horizontal circle of diameter 2 m, with zero angular velocity and a tangential acceleration given by " 4 t ". If the string breaks off at t = 5 s, then find the speed of the particle with which it flies off the circular path.

Solution : Here, tangential acceleration of the particle at time "t" is given as :

a T = 4 t

We note here that an expression of tangential acceleration (an higher attribute) is given and we are required to find lower order attribute i.e. linear speed. In order to find linear speed, we need to integrate the acceleration function :

a T = đ v đ t = 4 t

đ v = 4 t đ t

Integrating with appropriate limits, we have :

đ v = 4 t đ t = 4 t đ t

v f v i = 4 [ t 2 2 ] 0 5 = 4 X 5 2 2 = 50 m s

v f 0 = 50

v f = 50 m s

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Average total acceleration

Problem : A particle is executing circular motion. The velocity of the particle changes from (0.1 i + 0.2 j ) m/s to (0.5 i + 0.5 j ) m/s in a period of 1 second. Find the magnitude of average total acceleration.

Solution : The average total acceleration is :

a = Δ v Δ t = ( 0.5 i + 0.5 j ) - ( 0.1 i + 0.2 j ) 1

a = ( 0.4 i + 0.3 j )

The magnitude of acceleration is :

a = ( 0.4 2 + 0.3 2 ) = 0.25 = 0.5 m / s

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Problem : A particle starting with a speed “v” completes half circle in time “t” such that its speed at the end is again “v”. Find the magnitude of average total acceleration.

Solution : Average total acceleration is equal to the ratio of change in velocity and time interval.

a avg = v 2 v 1 Δ t

From the figure and as given in the question, it is clear that the velocity of the particle has same magnitude but opposite directions.

Circular motion

The speeds of the particle are same at two positions.

v 1 = v

v 2 = - v

Putting in the expression of average total acceleration, we have :

a avg = v 2 v 1 Δ t = v v t

a avg = 2 v t

The magnitude of the average acceleration is :

a avg = 2 v t

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Total acceleration

Problem : The angular position of a particle (in radian), on circular path of radius 0.5 m, is given by :

θ = - 0.2 t 2 - 0.04

At t = 1 s, find (i) angular velocity (ii) linear speed (iii) angular acceleration (iv) magnitude of tangential acceleration (v) magnitude of centripetal acceleration and (vi) magnitude of total acceleration.

Solution : Angular velocity is :

ω = đ θ đ t = đ đ t 0.2 t 2 0.04 = 0.2 X 2 t = 0.4 t

The angular speed, therefore, is dependent as it is a function in "t". At t = 1 s,

ω = 0.4 rad / s

The magnitude of linear velocity, at t = 1 s, is :

v = ω r = 0.4 X 0.5 = 0.2 m / s

Angular acceleration is :

α = đ ω đ t = đ đ t - 0.4 t = 0.4 rad / s 2

Clearly, angular acceleration is constant and is independent of time.

The magnitude of tangential acceleration is :

a T = α r = 0.4 X 0.5 = 0.2 m / s 2

Tangential acceleration is also constant and is independent of time.

The magnitude of centripetal acceleration is :

a R = ω v = 0.4 t X 0.2 t = 0.08 t 2

At t = 1 s,

a R = 0.08 m / s 2

The magnitude of total acceleration, at t =1 s, is :

a = a T 2 + a R 2

a = { 0.2 2 + 0.08 2 } = 0.215 m / s 2

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Problem : The speed (m/s) of a particle, along a circle of radius 4 m, is a function in time, "t" as :

v = t 2

Find the total acceleration of the particle at time, t = 2 s.

Solution : The tangential acceleration of the particle is obtained by differentiating the speed function with respect to time,

a T = đ v đ t = đ đ t t 2 = 2 t

The tangential acceleration at time, t = 2 s, therefore, is :

a T = 2 X 2 = 4 m / s 2

The radial acceleration of the particle is given as :

a R = v 2 r

In order to evaluate this expression, we need to know the velocity at the given time, t = 2 s :

v = t 2 = 2 2 = 4 m / s

Putting in the expression of radial acceleration, we have :

a R = v 2 r = 4 2 4 = 4 m / s 2

The total acceleration of the particle is :

a = a T 2 + a R 2 = 4 2 + 4 2 = 4 2 m s 2

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Questions & Answers

How can we take advantage of our knowledge about motion?
Kenneth Reply
pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
Mercy Reply
what are scalars
Abdhool Reply
show that 1w= 10^7ergs^-1
Lawrence Reply
what's lamin's theorems and it's mathematics representative
Yusuf Reply
if the wavelength is double,what is the frequency of the wave
Ekanem Reply
What are the system of units
Jonah Reply
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
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Amar
water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
sajjad
what is the definition of resolution of forces
Atinuke Reply

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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