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is ,  . An electron of mass kg revolves around a nucleus in a circular orbit of radius . Å. What is the angular momentum of the electron? (Velocity of electron is, v = . ms ) Solution: Mass of the electron, m = × − kg Radius of the electron, r = . Å = . m Velocity of the electron, v = . × ms - Angular momentum of electron is, L = I ω Electron is considered as a point mass. Hence, its moment of inertia is, I = m r The relation, v r could be used. Angular momentum, L = mr × v = mvr = . L = . × − kg m s − . A solid sphere of mass kg and radius . m rotates about an axis passing through the centre. What is the angular momentum if the angular velocity is rad s - Solution: Mass of the sphere, m = kg Radius r = . m - - - - Appendix Angular velocity ω = rad s - Solution: Angular momentum L = Iω = mr w = ( . ) ( . ) = . L = . kg m s - . A solid cylinder when dropped from a height of m acquires a velocity while reaching the ground. If the same cylinder is rolled down from the top of an inclined plane to reach the ground with same velocity, what must be the height of the inclined plane? Also compute the velocity. Solution: 2m h h ʹ In the first case, potential energy = kinetic energy mgh = mv mg× = mv ( ) In second case, potential energy = translational kinetic energy + rotational kinetic energy mghʹ = mv + I w mghʹ = mv + mr   ∴ mghʹ = mv ( ) Dividing ( ) by ( ), mgh mg mv mv ’ hʹ = m From equation ( ), mg mv g g ms . A small particle of mass m is projected with an initial velocity v at an angle θ with x axis in X-Y plane as shown in Figure. Find the angular momentum of the particle. - - - - Appendix Solution: v cos θ θ v sin θ Let the particle of mass m cross a horizontal distance x in time t. Angular momentum L dt But  xi yj and mgj = −