Question

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plz solve this question.


There is a hemispherical surface rotating about an Axis as shown in the figure. What will be the angular speed ? ( constant) , so that the block and stay as in the figure , considered friction to be absent .


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Answer 1

Answer:

→ $\boxed{ \sf \omega = \sqrt{ \frac{5g}{3R} .}}$

Explanation:

See the attachment.

Hence it is solved..

Answer 2

Answer:

$\boxed{ \pink{ \tt \therefore \omega = \sqrt{ \frac{5g}{3N \: R} } .}}$

Explanation:

We have,

Angel = 53° .

$\tt \sin 53 = \frac{r}{R} \\ \\ \tt \implies \frac{4}{5} = \frac{r}{R} \\ \\ \tt \therefore \: r = \frac{4R}{5} .$

First see, in vertical direction,

$\sf N \: cos53 \: = mg. \\ \\ \sf \implies N \times \frac{3}{5} = mg. \\ \\ \sf \implies N = \frac{5}{3} mg..........(1).$

Now, in horizontal direction :

$\sf\because N \: sin53 = ma_c. \\ \\ \sf \implies \frac{ \cancel5 \cancel{m}g}{3} \times \frac{4}{ \cancel5} = \cancel{m} { \omega}^{2} r. \: \: \: \: \{from \: (1) \} \\ \\ \sf \implies \frac{ \cancel4g}{3} = \frac{ { \omega}^{2} \cancel4 R}{5} . \\ \\ \sf \implies { \omega}^{2} = \frac{5g}{3R} . \\ \\ \boxed{ \pink{ \tt \therefore \omega = \sqrt{ \frac{5g}{3N \: R} } .}}$

Hence, it is solved.