Respuesta :
Answer:
Time taken, [tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]
Explanation:
It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.
From the figure,
The sum of forces in y direction is :
[tex]T\ cos\theta-mg=0[/tex]
[tex]T=\dfrac{mg}{cos\theta}[/tex]
Sum of forces in x direction,
[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]
[tex]mg\ tan\theta=\dfrac{mv^2}{r}[/tex].............(1)
Also, [tex]r=l\ sin\theta[/tex]
Equation (1) becomes :
[tex]mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}[/tex]
[tex]v=\sqrt{gl\ tan\theta.sin\theta}[/tex]...............(2)
Let t is the time taken for the ball to rotate once around the axis. It is given by :
[tex]T=\dfrac{2\pi r}{v}[/tex]
Put the value of T from equation (2) to the above expression:
[tex]T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]
[tex]T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]
On solving above equation :
[tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]
Hence, this is the required solution.
The time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].
How long does it take for the ball to rotate once around the axis?
As it is given to us that the small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone.
Now since the ball is moving in circular as well as vertical motion, therefore, the sum of vertical forces at any given moment of time can be written as,
[tex]\rm T\ cos \theta = mg\\\\T = \dfrac{mg}{Cos\theta}[/tex]
Also, the sum of the forces in the x-direction,
[tex]\rm T\ sin\theta= \dfrac{mv^2}{r}[/tex]
Substitute the value of T,
[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{r}[/tex]
[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{l\ sin\theta}[/tex]
[tex]v=\sqrt{gl\ tan\theta \cdot sin\theta}[/tex]
We know that the time is taken by the ball to circulate around the axis can be given by the formula,
[tex]T = \dfrac{2\pi r}{v}[/tex]
Substitute the value of v,
[tex]T = \dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta\cdot sin\theta}}[/tex]
[tex]T=2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex]
Hence, the time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].
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