Question

Semi-vertical angle of the conical section of a funnel is 37*(degree). there is a small ball kept inside the funnel. on rotating the funnel. on rotating the funnel, the maximum speed that the ball can have in order to remain in the funnel is 2 m/s. calculate inner radius of the brim of the funnel. Is there any limit upon the frequency of rotation? how much is it? Is it lower or upper limit? Give a logical reasoning. (use g=10m/s square and sin 37*(degree)=0.6)
The image is solution of the above question.

ACTIVITY
using a funnel and a marble or ball bearing try to work out the situation in the above question try to realize that as the marble goes towards the brim, its linear speed increases but it's angular speed decreases. when nearing the base, it is the other way.
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Answer 1

Answer:

ANSWER

From the figure,

Rsin θ=mg

and, Rsin θ=mv

2

/r

⟹tan θ=gr/v

2

=r/h

⟹v=

gh

=

10×0.1

=1m/s

Option D is correct

Answer 2

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Question

semi - vertical angle of the conical section of a funnel is 37°.There is a small ball kept inside the funnel.on rotating the funnel,the maximum speed that the ball can have in order to remain in the funnel is 2 m/s.Calculate the inner radius of the brim of the funnel.Is there any limit upon the frequency oh rotation?How much is it ?It is lower or upper limit ?Give a logical reasoning.(Use g = 10 m/s² and and sin 37° = 0.6)

Solution

N Sinθ = mg and N cosθ = mv²/r

∴tan θ = rg/v² .°. r = v² tanθ/g

∴ ${\sf{ \large{r}}}$${ \sf{ \small{max}}}$$\sf{ { = v}^{2}}$${ \sf{ \small{max}}}{ \sf{ \frac{tanθ}{g} = 0.3m}}$=v²max tanθ=0.3m

v = rw = 2π rn

If we go for the lowest limit of the speed(while rotating), v → 0 .°. r →0,but the frequency n increases.

Hence a specific upper limit is not possible in the case of frequency.

Thus,the Practical limit on the frequency of rotation is it lower limit.It will be possible for r = r^max

∴ n = n ^max /2πr ^max= 1/0.3π = 1 rev / s.

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