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Motion Circular.

posted Aug 12, 2015, 9:00 PM by Upali Salpadoru   [ updated May 12, 2016, 12:50 PM ]

Moon orbiting the Earth, the planets and a car taking a bend are all instances of circular motion.


Cir m 1.jpg

Fig. 1. A fast bowler in action.

The picture illustrates a bowler in action. His hand takes almost a semicircle before delivery. The moment he releases the ball it moves at a tangent to the circle.


Next picture shows a thrower in action. He rotates the ball a few times in a horizontal plane and releases it. The ball flies making a 90° angle to the radius.

C M 2.jpg

As long as the object is in uniform circular motion there will be an un -balanced force acting on it towards the centre. This is the centripetal force. This can be calculated by the formula given below.





Fig. 2 Rotating a heavy mass.


Finding the speed of a revolving object.

In order to get the speed we must know the distance and the time.  The time taken for one complete circle is called the Period. (symbol T)The circumference of the circle will be the distance. The circumference is = 2µr.

Velocity =  2µr / T

In symbols they are expressed as follows:-

 



Before starting to calculate forces it is necessary to revise trigonometrical ratios. At least you will have to learn Resolution of Vectors.

If a vector is represented by R it can be resolved in vertical and horizontal directions using this diagram.




Resolving.jpg

Vertical component = R cos θ.

 Horizontal component  = R sine θ.



  Fig.3 Resolving vectors.













Fig 3. Connecting weight, centripetal force and tension.

C M 3.jpg


If the mass of the object is 3.0 kg let us find out the forces now.

Assuming g to be 9.8 Nkg-1the weight of the object will be Mg =  N.

As the angle between vertical and the rope is 60˚   

Ft cos 60 ˚ =  Mg

Ft = 3x 9.8  / cos 60 ˚   ( cos 60 = 0.5)

Ft =29.4   x 0.5 = 14.7 N. ……..This is the tension force.



14.7  sine 60˚ =  F c.

Fc =14.7 x  0. 87 N .

F c =   12.8  N      …………….This is the centripetal force.

VERTICAL  CIRCULAR  MOTION

  Tie a can of water to a thread and turn it round gradually changing it to a vertical plane as shown in the diagram. If you can turn it with a sufficient speed water will not come down. What is this sufficient speed?

As the can goes round and round The thread keep on pulling the water can. This is tension on the string and provides the centripetal force. (Fc) .  V m 1.jpg

As we have learned earlier Fc = mv2 / r

The length of the string is 1m.

If the water can has a mass of 0.5 kg and does 10 rounds in one second we can find the Fc.

In order to maintain vertical motion , the minimum centripetal force required is equal to the weight.


mg = Fc

m x g = m xv2 / r  =  

As ‘m’ cancels    gx r  = v2

Velocity =  Distace / time

             = 10 x 2π r/10

                 = 2π r

                 = 2 x 3.14 x1 = 6.28 ms-1

                             Fc = 0.5 x 6.28 2   /`1     = 19.7 N

As this is greater than weight vertical motion is possible..


Motion in a circle.

The groove in the record had to pass the needle at a specific speed for accurate rendition of music. The earliest records had a rotatory speed of 78 RPM.  Does this mean that the needle meets the record at a constant speed?  No! not at all. As the needle approaches the centre the speed at which groove passes the needle changes. Let us do a calculation and examine how it happens.

 Fig.1. The record does 78 revolutions per minute.

Let us assume that the needle is at the outermost point.
As the record turns 78 times in a minute , the time it takes for one turn is :-  
1/78 minutes (ie. 0.77seconds)  This time is known as the period and denoted by the letter T.  

 
The radius of a record is 15 cm; which is  0.15 m.   

From this we can calculate the circumference using 2πr

The circumference of the record  :-  2x 3.14x 0.15  =  0.94 m.
Therefore the speed of a point  at the circumference  = d/t  

                                             =   0.94 / 0.77 =    1.22 ms-1


As the record is played, the needle gets closer and closer to the centre. If it would be about 5cm, that is 0,05 m away from the centre, What would be the speed for the needle?.
The circumference = 2 πr  = 2x 3.14x0.05m = 0.314 m.
Then the speed would be  =   0.314 /  0.77 ms-1  (Time for a turn remains unchanged)  
                                         =0.41ms-1


Fig. 2. Rotating a fire ball at a uniform speed.


The fig. shows a man rotating a fire ball. The length of the chain is 1 m. Mass of the ball is 250 g (0.25kg).If he does 60 rpm,  he’ll be taking 1 second for a single turn. The period  is 1 s.    

What would be the speed of the ball?


Speed = distance / time  =   2πr/1  ms-1.    =  (2x3.14 x1) / 1  = 6.28 ms-1
Acceleration  =  v2/r       =  (6.28 x 6.28) / r      =    39.4 ms-2
Centripetal force  = mass x acceleration   =  0.25 x 39.4 =  9.9 N to the centre.


Easy formula for acceleration

 

Velocity =  Circumference  / T   =  2 πr  /T

Acceleration      =     v2/r    =  (2 πr/T)2/r

                          =     π2r /  T2                      

                          =     4  x 9.8 x 1/1

                          =    39.4ms-2

 

 

 



Question. 1.0   

The man is turning a mass tied to a string as shown. Length of the string is 1.0 m. The object does 16 swirls in 5 seconds.Q  1.jpg


Find the following:-

1.1  What is the period ‘T’ of circling?

1.2  If the thread snaps at the point D what will be the velocity of object leaving the circle ?

1.3  What is the acceleration?

1.4  If the mass of object is 400g (0.4kg)  what was the tension on the string?





Question 2.0


Q 2.jpg

A red disc of mass 200g is placed at the circumference of aa rotating table. The limiting friction between the disc and the table is 3 N.

2.1 What is the minimum speed necessary for it to get thrown out.

2.2 What is the number of revolutions per minute (rpm) at the time?







Question 3.0


Q 3.4.jpg

Q 3.2.jpg



A car having a mass of 800 kg is on a bend banked at 20˚. The radius of the curve is 50 m.

Find the following:-

3.1 Vertical force on the car.

3.2 Force acting normal to the road.  (N)

3.3 Horizontal component due to weight.

3.4 Maximum speed it can use without any friction on tyres.





For Answers Click :- Answer page A to K.

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