Remove th refill from a ball point pen. Pass an about 1m long thin, strong cottonthread through the body of the pen. Tie a heavy washer A at the lower end and acomparatively lighter washer B at the upper end as shown in fig. The masses ofwashers must be precisely measured with the help of a physical balance. Mark a pointon the thread at a distance of 20 cm below the lower edge of the body of the pen. Holdthe pen vertically in your right hand and rotate the washer B till it moves on a circularpath in horizontal plane with constant speed. You will find that the washer A movesup. Increase the speed of rotation till the mark on the thread rises to the lower edge ofthe body of the pen. Calculate the speed with which washer B is moving on its circu-lar path. Repeat the experiment for thread lengths 30 cm, 40 cm and 50 cm below thelower edge of the pipe. Plot a graph between v2 and r taking v2 on y - axis and r on x-axis
Answers
Answer:
Explanation:
Remove the refill from a ball point pen then pass a long and strong cotton thread through the body of the pen. Tie a heavy washer A at the lower end and a comparatively lighter washer B at the upper end. Hold the pen vertically in your right hand. It is moving on it's circular path.
We have to plot a graph between V2 and r taking v2 on y-axis and r on x-axis.
Consider a particle of mass on moving with a uniform speed v in a circle. After a small time ∆t, the particle reached B and its velocity is represented by the tangent B directed along by r and r' be the position vectors and V and Vi the velocities of particle at A and B respectively as shown in fig.
To determine the change in velocity ∆v due to the change in direction consider a point O outside. The circle fig (a) and V1 and r, r' in each other
:• Acceleration = Rate of change of velocity
a = PQ/∆t = ∆v/∆t
Where acceleration 1ar to ∆r
•: |r| = |v'| op = OQ - OP in fig
As t is very small, AB is also very small and is nearly a straight line. Then triangle ABC and triangle POQ are isosceles triangles having their included angles equal. The triangle are therefore, similar and hence,
PQ/AB = OP/CA
or ∆v/V∆t = v/r
[As magnitudes of velocity vectors V1 and V2 = V (say) ]
or ∆v/∆t = v2/r
But ∆v/∆t is the acceleration of the particle . Hence, centripetal acceleration, a = v2/r
Since v = rw, the magnitude of centripetal force is given by
F = ma = mv2/r = mrw2
As ∆t is very small, ∆theta is also very small and angle OPQ = angle OQP = 1 right angle.
Thus PQ is perpendicular to OP, which is parallel to the tangent AX at A. Now AC is also perpendicular to AX. Therefore AC is parallel to PQ. It shows that the centripetal force at any point acts towards the centre along the radius.
It shows that some minimum centripetal force has to be applied on a body to make it move in a circular path. In the absence of such a force, the body will move in a straight line path.
