Question

A cubical block of side a is moving with velocity v on a horizontal smooth plane, as shown it hits a ridge at point O.
The angular speed of the block after it hits O is.

Ans. 3v/4a.
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☆ Help me solve this ques.
☆☆ if you're applying momentum conservation, then pls explain here, Why energy conservation is not applied??
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Answer 1

4) zero is the right answer ok
Mark me as brainlist plz plz plz

Answer 2

Given:

A cubical block of side a is moving with velocity v on a horizontal smooth plane, as shown it hits a ridge at point O.

To find:

Angular speed after hitting ridge O ?

Calculation:

• One thing can be clearly understood that, after hitting ridge O , the block will topple.

• Now, while toppling, the block doesn't experience any external torque.

• So, the angular momentum will remain conserved.

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• Another important thing is calculation of Moment of Inertia of the block during toppling (see diagram)

So, applying Angular Momentum Conservation:

$\therefore \: L_{1} = L_{2}$

$\implies \: m \times v \times r = I \times \omega$

$\implies \: m \times v \times r = \bigg \{\dfrac{m {a}^{2} }{6} + m {( \dfrac{a}{ \sqrt{2} }) }^{2} \bigg \} \times \omega$

$\implies \: m \times v \times \dfrac{a}{2} = \bigg \{\dfrac{m {a}^{2} }{6} + m {( \dfrac{a}{ \sqrt{2} }) }^{2} \bigg \} \times \omega$

$\implies \: m \times v \times \dfrac{a}{2} = \bigg \{\dfrac{m {a}^{2} }{6} + \dfrac{m {a}^{2} }{2} \bigg \} \times \omega$

$\implies \: m \times v \times \dfrac{a}{2} = \bigg \{\dfrac{(3 + 1)m {a}^{2} }{6}\bigg \} \times \omega$

$\implies \: m \times v \times \dfrac{a}{2} = \bigg \{\dfrac{4m {a}^{2} }{6}\bigg \} \times \omega$

$\implies \: m \times v \times \dfrac{a}{2} = \bigg \{\dfrac{2m {a}^{2} }{3}\bigg \} \times \omega$

$\implies \: v \times \dfrac{a}{2} = \bigg \{\dfrac{2 {a}^{2} }{3}\bigg \} \times \omega$

$\implies \: v \times \dfrac{1}{2} = \bigg \{\dfrac{2 a }{3}\bigg \} \times \omega$

$\implies \: \omega = \dfrac{3v}{4a}$

Hope It Helps.