Physics 9th: Motion | Circular Motion and Some Solved Examples on Motion – Time Graph

Physics 9th: Motion | Circular Motion and Some Solved Examples on Motion – Time Graph

 

Circular Motion

 

    When a body or an object moves in a circle, its motion is called circular motion. In other words, motion in a circle is circular motion.

    When a body (or object) is moving along a circular path, then its direction of motion (or direction of speed) keeps changing continuously.

    Since the velocity changes (due to continuous change in direction), therefore, the motion along a circular path is said to be accelerated.

    When a body moves in a circular path with uniform speed (constant speed), its motion is called uniform circular motion.

    Figure (a) shows the path of an athlete along a rectangular track PQRS. The athlete changes his direction of motion at the four corners P, Q, R, S of the rectangular track, in every round.

Figure (b) shows a hexagonal path where in every round, the direction of motion is changed at the six corners P, Q, R, S, T, U and P.

In figure (c) the track is a regular octagon. In every round, the athlete has to change the direction of motion at eight corners P, Q, R, S, T, U, V, W of the octagon; and so on. We observe that as number of sides of track increase, athlete has to take turn more and more number of times.

If we go on increasing the number of sides of the track indefinitely, we find that the shape of the track approaches the shape of a circle. The length of each side tends to be zero.

Along the circle, the athlete has to change his direction of motion at each point.

Therefore, the effective direction of motion is along the tangent to the circular path at that point as shown in figure (d).

 

    The force which is needed to make an object travel in a circular path is called centripetal force.

    Artificial satellites move under uniform circular motion around the earth.

    The earth moves around the sun in uniform circular motion

    The tip of a seconds’ hand of a watch exhibits uniform circular motion on the circular dial of the watch.

    \displaystyle \text{speed}=\frac{\text{Distance}}{\text{Time}}    \displaystyle v=\frac{2\pi r}{t}

        where        v = speed

        (pi)        \displaystyle \pi =\frac{22}{7} (It is a constant)

                r = radius of circular path

        and        t = time taken for one round of circular path

 

Some Solve Examples Based on Motion – time Graphs

 

Question:     Which graph represents the case of:

            (i)    A cricket ball thrown vertically upwards and returning to the person

            (ii)    A trolley decelerating to a constant speed and then accelerating

 

Solution:     

Graph C represent cricket ball moving vertically up and then coming down. This is because in this case speed first decreases and when the ball starts coming down, speed increases.

        Graph A represents a trolley which decelerates, remains at constant speed and then accelerates.

 

Question:     In the figure two displacement–time graphs have been drawn. Which represents a higher velocity and why?

Solution:     

B represents higher velocity because slope of B is more than that of A.

 

Question:     In figure two velocity-time graphs have been shown. Which represents higher acceleration? Why?

Solution:

Acceleration ion of both is same because two graphs are parallel lines and their slopes are equal.

 

Question:     Is speed-time graph shown in figure possible? Why?

Solution:     

Not possible because speed can never be negative.

 

Question:     Figure shows displacement-time curves of two particles I and II. What conclusions do you draw from these curves?

 

Solution:     (i) Both the curves are representing uniform linear motions.

        (ii) Initial value of displacement for II is zero but has a finite positive value of I

        (iii) Uniform velocity of II is more than that of I because slope of curve II is greater.

 

Question:     In figure five distance-time graphs have been drawn. What type of motion do these graphs represent?

(i) 

(ii) 

(iii) 

(iv) 

(v) 

 

Solution:     

(i) Particle is at rest and no motion is taking place.

    (ii) Uniform linear motion

    (iii) And (iv) Non-uniform motion

    (v) Uniform linear motion    

 

Question:     Does the graph shown in figure represent uniform motion?

Solution:     (i) No, it does not represent uniform motion.

 

Question:     The speed-time graph for a car is shown in figure.

    (i) Find how does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

    (ii) Which part of the graph represents uniform motion of the car?

 

Solution:     

        (i) From the speed time graph shown in figure we find that speed of the body goes on increasing from t = 0 to t = 6 s. As the speed time graph is not a straight line, therefore, acceleration of the body is not uniform.

    The distance travelled by the car in the first four second is given by the area enclosed by the speed time graph and x-axis from t = 0 to t = 4s. This has been lightly shaded in the graph.

    To calculate the distance, in such cases, we actually count the number of squares in the shaded portion of the graph. It comes out to be 320.5 square. As one square on X-axis represents \displaystyle t=\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}m.

    \ Total area of 320.5 squares \displaystyle =320.5\times \frac{1}{18}m = 17.80 m

    (ii) In uniform motion, speed of the body becomes constant. The portion of the        graph from t = 6s to t = 10 s represents uniform motion of the car.

 

Question:     An artificial satellite is moving in a circular orbit of radius 42, 250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

 

Solution:     

Here,     r = 42250 km, t = 24 h

    speed    v = ?

    from    \displaystyle v=\frac{2\pi r}{t}

        \displaystyle v=2\frac{22}{7}\times \frac{42250}{24}=1.106\times {{10}^{4}} km/h

        \displaystyle \frac{1.106\times {{10}^{4}}}{60\times 60s}=3.07 km/s

 

Question:     The velocity-time graph of an object under linear motion is shown in figure. What is its acceleration? Find the total distance covered by the object in 8 s

Solution:     

Acceleration of the object = slope of \displaystyle v-t graph

        \displaystyle \frac{-20m{{s}^{-1}}}{8s}=-2.5m{{s}^{-2}}

    As the graph is sloping downward, hence the slope has been taken negative.

    Total distance covered by the object in 8 s = area of triangle formed by \displaystyle v-t curve

        \displaystyle =\frac{1}{2}\times base\times height

        \displaystyle =\frac{1}{2}\times \left( 8s \right)\times \left( 20\,m{{s}^{-1}} \right)=80 m

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Physics 9th: Motion | Derivation of Equations of Motion by Graphical Method

Physics 9th: Motion | Derivation of Equations of Motion by Graphical Method

Derivation of Equations of Motion by Graphical Method

 

TO DERIVE v = u + at BY GRAPHICAL METHOD

 

This is a graph of uniform acceleration with ‘u’ as initial velocity and ‘v’ as final velocity.

    Initial velocity = u = OP

    Final velocity = v = RN

             = RQ + QN

             v = u + QN    …..(i)

    Acceleration, a = slope of line PN

         \displaystyle a=\frac{QN}{PQ}=\frac{QN}{t}

         \displaystyle QN=at        …..(ii)

    Putting the value of QN from equation (ii) into equation (i), we get v = u + at        

 

TO DERIVE \displaystyle s=ut+\frac{1}{2}a{{t}^{2}} BY GRAPHICAL METHOD

 

    In the above speed-time graph, the distance travelled is given by

            Distance travelled = Area of figure OPNR

                     = Area of DPNQ + Area of rectangle OPQR

    (1)     Area of triangle PNQ = \displaystyle \frac{1}{2}\times base\times height

                     = \displaystyle \frac{1}{2}\times PQ\times NQ \displaystyle =\frac{1}{2}\times t\times (v-u)

                     = \displaystyle \frac{1}{2}\times t\times at    [As v = u + at and v u = at]

             Area of DPNQ = \displaystyle \frac{1}{2}a{{t}^{2}}

    (2)    Area of rectangle OPQR = OP ´ PQ

                      = u ´ t

                      = ut    

    Distance travelled = Area of DPNQ + Area of rectangle OPQR

    \displaystyle S=\frac{1}{2}a{{t}^{2}}+ut

    \displaystyle S=ut+\frac{1}{2}a{{t}^{2}}

TO DERIVE \displaystyle {{v}^{2}}-{{u}^{2}}=2aS BY GRAPHICAL METHOD

 

    In the above speed time graph distance travelled (S) = Area of trapezium OPNR

    \displaystyle S=\frac{1}{2}\times (sum of parallel sides) ´ height

    \displaystyle S=\frac{1}{2}\times (OP + RN) ´ OR

    \displaystyle S=\frac{1}{2}\times (u + v) ´ t

    \displaystyle S=\frac{1}{2}\times (v + u) t            …(i)

    But v = u + at

    at = v u

    \displaystyle t=\left( \frac{v-u}{a} \right)                … (ii)

    Putting this value of ‘t’ from equation (ii) into equation (i) we get that ___

    \displaystyle S=\frac{1}{2}\times (v + u) \displaystyle \left( \frac{v-u}{a} \right)

    \displaystyle S=\frac{\left( v+u \right)\left( v-u \right)}{2a}

    \displaystyle S=\frac{{{v}^{2}}-{{u}^{2}}}{2a}        [As (a + b) (a b) = \displaystyle {{a}^{2}}-{{b}^{2}}]

    2as = \displaystyle {{v}^{2}}-{{u}^{2}}

    \displaystyle {{v}^{2}}-{{u}^{2}}=2as

        

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