Question

A hemispherical bowl of radius R=0.1m is rotating about its own axis (which is verticle) with an angular velocity omega. A particle of mass 10^(-2)kg on the smooth inner surface of the bowl is also rotating with the same omega. The particle is at a height h from the bottom of the bowl (a) obtain the relation betweemn h and omega. what is the minimum value of omega needed, in order to have a non-zero value of h? (b) it is desired to measure g using this set up, by measuring h accurately. assuming that R and Omega are known precisely and least count in the measurement of h is 10^(-4)m, what is the minimum possible error Deltag in the measured value of g? (g=10m//s^(2))

Answer 1

Answer:

Part a)

Relation in height h and angular speed is given as

$R - \frac{g}{\omega^2} = h$

also for minimum value of angular speed

$\omega = \sqrt{\frac{g}{R}}$

Part b)

Minimum possible error in g is

$\Delta g = 10^{-4} m/s^2$

Explanation:

Let the height of the bowl is given as "h"

So here we can say that the centripetal force is given by normal force due to bowl

So we have

$N cos\theta = mg$

$N sin\theta = m\omega^2 r$

$tan\theta =\frac{\omega^2 r}{g}$

$tan\theta = \frac{\omega^2 R sin\theta}{g}$

$cos\theta = \frac{g}{\omega^2 R}$

also we know that

$R - Rcos\theta = h$

$R - \frac{g}{\omega^2} = h$

also for minimum value of angular speed

$\omega = \sqrt{\frac{g}{R}}$

Part b)

From above relation we know that

$R - \frac{g}{\omega^2} = h$

since angular speed and radius is measured precisely

so we have

$\Delta h = \Delta g$

$\Delta g = 10^{-4} m/s^2$

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Topic : Circular motion

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Answer 2

Answer:

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