A body of mass m is projected at an angle 45 degree with horizontal with velocity u. Find an expression for angular momentum of the body at its maximum height of projection
Answer is (m*u^3)/(4*sqrt (2)* g)
Answer is (m*u^3)/(4*sqrt (2)* g)
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at max , height usin(45) is zero . so only ucos(45) is left
L = p*r
= m*V*H
V = ucos45 = u/sqrt(2)
H = u^2sin^2(theta)/2g (formula)
= u^2sin^2(45)/2g
= u^2/4g
L = m*u/sqrt(2) * u^2/4g
= mu^3/(4*sqrt(2)*g)
L = p*r
= m*V*H
V = ucos45 = u/sqrt(2)
H = u^2sin^2(theta)/2g (formula)
= u^2sin^2(45)/2g
= u^2/4g
L = m*u/sqrt(2) * u^2/4g
= mu^3/(4*sqrt(2)*g)
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At max height H = Uy^2/2g, the angular momentum is L = Iw = kmH^2 * Ux/H = mUxH = mUxUy^2/2g = m U^3 cos(45) sin(45)^2/2g = m U^3 sqrt(2)/2 sqrt(2)^2/4 // 2g = m U^3 sqrt(2) 2 //(8 2 g) = m U^3/(4 sqrt(2) g) QED.
Frankly, it's not at all clear why anyone would want the answer in the given format. m U^3 sin(90)/4g = m U^3/4g is much more interesting and useful. NOTE 2 cos(theta) sin(theta) = sin(2theta).
Frankly, it's not at all clear why anyone would want the answer in the given format. m U^3 sin(90)/4g = m U^3/4g is much more interesting and useful. NOTE 2 cos(theta) sin(theta) = sin(2theta).