Question

3. In a conical pendulum the height of point of suspension from the horizontal plane of circle is 9.8 m/ g= 9.8 m/s square) the period of conical pendulum----​

Answer 1

Answer:

.E.F image

In vertical direction,

Tsinθ=mg(i)

In horizontal plane,

Tcosθ=

R

mv

2

[centripetal for

From (i), T=

sinθ

mg

=

l

2

−R

2

mgl

T=(

1000

150

)

(120)

2

−(20)

2

(10)×120

=1.52N

Answer 2

Answer:

The time period of conical pendulum will be 6.28s.

Explanation:

The pendulum will make a certain angle say'θ' at the point of suspension.

There will be two kinds of tension acting on the conical pendulum, the first one will be in the vertical plane and the second one will be in the horizontal plane.

The vertical tension will be balanced by its weight i.e.

Tsinθ=mg   (1)

Tsinθ=tension in the vertical plane

m=mass of the pendulum

g=acceleration due to gravity

and the tension in the horizontal direction will be balanced by centripetal force,

Tcosθ=$mV^{2} /R$     (2)

Tcosθ=tension in the horizontal plane

m=mass of the pendulum

V=velocity

R=radius of the pendulum

By taking the ratio of equations (1) and (2) we get;

tanθ=$gR/V^{2}$     (3)

The relation between velocity and angular acceleration;

V=ω×R         (4)

By placing equation (4) in equation (3) we get;

tanθ=g$/$$($ω$)^{2}$R

The height of  the pendulum will be equal to the radius,

Also, when the vertical and horizontal force is balanced then the angle of suspension will be'θ=45°',

1=g$/$$($ω$)^{2}$R

$($ω$)^{2}$=g/R

ω=$\sqrt{g/R}$. (5)      

values given in question are,

g=9.8$ms^{-2}$

R=9.8m

By putting them in equation 5 we get;

ω=1

also,

ω=2π$/$T

T=time period

1=2π$/$T

T=2π×1=2×3.14=6.28 seconds

Hence, the time period of conical pendulum will be 6.28s.