Question

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Answer 1

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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}}$

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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Answer 2

Answer:

N Sinθ = mg and N cosθ = mv²/r

$∴tan θ = rg/v² .°. r = v² tanθ/g$

${\sf{ \large{r}}}{ \sf{ \small{max}}}rmax \sf{ { = v}^{2}}{ \sf{ \small{max}}}{ \sf{ \frac{tanθ}{g} = 0.3m}}=v </p><p>2</p><p> max </p><p>g</p><p>tanθ</p><p> </p><p> =0.3m$

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

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

Explanation:

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