Question

A large platform is moving with constant acceleration 'a' perpendicular to its plane in gravity free space. A particle of mass m is projected with speed u relative to the platform at an angle theta with its plane from a point O on it. The angular momentum of the particle about O:

Answer 1

Answer:

Angular momentum of the particle is given as

$L = m(\frac{1}{2} at^2 cos\theta)$

Explanation:

As we know that the object is projected with speed "u"

So we have

acceleration of the object in vertical direction is given as "a"

Now we have

$v_y = u sin\theta - at$

$v_x = ucos\theta$

now the distance along x and y direction traveled by it is given as

$x = u cos\theta t$

$y = u sin\theta  t - \frac{1}{2}at^2$

now angular momentum is given as

$L = mv_x y - m v_y x$

$L = m(ucos\theta)(u sin\theta t - \frac{1}{2}at^2) - m(usin\theta - at)(ucos\theta t)$

$L = m(u^2cos\theta sin\theta t - u\frac{1}{2} at^2 cos\theta) - m(u^2 sin\theta cos\theta t- ucos\theta at^2)$

$L = m(\frac{1}{2} at^2 cos\theta)$

#Learn

Topic : Angular momentum

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