Advertisements
Advertisements
प्रश्न
Draw a neat labelled diagram of conical pendulum. State the expression for its periodic time in terms of length.
Advertisements
उत्तर
Where, S: rigid support
T : tension in the string
l : length of string
h : height of support from bob
v : velocity of bob
r : radius of horizontal circle
θ: semi-vertical angle
mg : weight of bob
i) Consider a bob of mass m tied to one
end of a string of length ‘l’ and other
the end is fixed to a rigid support.
ii) Let the bob be displaced from its mean
position and whirled around a
horizontal circle of radius ‘r’ with
constant angular velocity ω, then the
bob performs U.C.M.
iii) During the motion, a string is inclined to
the vertical at an angle θ as shown in
the above figure.
iv) In the displaced position P, there are two
forces acting on the bob.
a. The weight mg acting vertically
downwards.
b. The tension T acting upward
along the string.
v) The tension (T) acting in the string can
be resolved into two components:
a. T cos θ acting vertically upwards.
b. T sin θ acting horizontally towards
centre of the circle.
vi) Vertical component T cos θ balances the
weight and horizontal component T sin θ
provides the necessary centripetal force.
∴ T cos θ = mg . ........(1)
T sin θ = `"mv"^2/"r" = "mr"omega^2` ....(2)
vii) Dividing equation (2) by (1),
tan θ = `"v"^2/"rg"` ......(3)
Therefore, the angle made by the string with the vertical is θ = `tan^-1 ("v"^2/"rg")`
Also, from equation (3),
v2 = rg tan θ
∴ v = `sqrt ("rg" tan theta)`
The period, T = `(2pi"r")/"v" = (2pi"r")/sqrt("rg" "tan" theta) = 2pi sqrt ("r"/("g tan" theta))`
It can be seen that r - = l sin θ
T = `2pi sqrt (("l sin" theta)/("g tan" theta)) = 2pi sqrt (("l cos" theta)/"g")`
The period of a conical pendulum is `2pi sqrt (("l cos" theta)/"g")`
APPEARS IN
संबंधित प्रश्न
For a particle performing uniform circular motion `vecv=vecomegaxxvecr`obtain an expression for linear acceleration of the particle performing non-uniform circular motion.
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Is it possible to have an accelerated motion with a constant speed? Explain
Earth moves around the sun with uniform velocity.
Give an example of motion in which speed remains uniform, but the velocity changes.
A uniform circular motion is an accelerated motion. Explain it. State whether the acceleration is uniform or variable? Name the force responsible to cause this acceleration. What is the direction of force at any instant? Draw a diagram in support of your answer.
A uniform linear motion is unaccelerated, while a uniform circular motion is an accelerated motion.
Complete a sentence and explain it.
When an object is in uniform circular motion, its ______ changes at every point.
Which of the following remains constant in a uniform circular motion, Speed or Velocity, or both?
Is it possible to have an accelerated motion with a constant speed? Explain.
If a particle moves with uniform speed then its tangential acceleration will be ______.
A particle goes round a circular path with uniform speed v. After describing half the circle, what is the change in its centripetal acceleration?
The ratio of angular speed of a hour-hand to the second-hand of a watch is ____________.
The ratio of the angular speed of minute hand and hour hand of a watch is ____________.
A particle performs uniform circular motion in a horizontal plane. The radius of the circle is 8 cm. The centripetal force acting on the particle is 15 N. Its kinetic energy is ____________.
A body of mass ·m' is moving along a circle of radius 'r' with linear speed 'v'. Now, to change the linear speed to `V/2` and to move it along the circle of radius '4r', required change in the centripetal force of the body is ______.
A stone of mass 3 kg attached at one end of a 2m long string is whirled in horizontal circle. The string makes an angle of 45° with the vertical then the centripetal force acting on the string is ______.
(g = 10 m/s2 , tan 45° = 1)
A cyclist is riding with a speed of 43.2 km/h. As he approaches a circular turn on the road of radius 60 m, he applies brakes and reduces his speed at constant rate of 1.8 ms-2. The magnitude of the net acceleration of the cyclist is ______.
The given graph represents motion with ______ speed.
A body moves in a uniform circular motion ______.
A body is said to be in nonuniform motion if it travels ______.
It is possible to have objects moving with uniform velocity but non-uniform acceleration.
A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is ______.
Two bodies of masses 10 kg and 5 kg moving in concentric orbits of radii R and r such that their periods are the same. Then the ratio between their centripetal accelerations is ______.
Explain the meaning of uniform circular motion.
What is the direction of the velocity of an object at any point during uniform circular motion?
