Question

2:41 session.masteringphysics.com AA Problem 7.42 MasteringPhysics 9 of 9 Constants A pendulum consists of a mass M hanging at the bottom end of a massless rod of length L, which has a frictionless pivot at its top end. A mass m, moving as shown in L M QTap image to zoom with velocity v, impacts M and becomes embedded. Part A What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear over the top of its arc? Express your answer in terms of g and some or all of the variables m, M, L.

Answer 1

the conservation of energy,

1/2​m*v^2 = m*g*h

The initial kinetic energy of the embedded system must be equal to the final potential energy of the system.

The total mass will be ( {m + M} and the height at the top is 2L

Substitute m+M for m and 2L for h in the above equation.

1​/2(m+M)v^2=(m+M)g(2L)

V^2 = 4gL

$\small v = \sqrt{4gL}$ ........1

The angular momentum of a system is,

L = mvr

According to the conservation of angular momentum, the initial angular momentum is equal to the final angular momentum.

For initial angular momentum, the mass is m , the velocity is v _{sufficient​} and distance is L ,

For final angular momentum, the mass is (m+M) , the velocity is v and distance is L .

m*v_sufficient​*L = (m+M)*v*L

v_sufficient = (m+M)*v / m

form eq. 1

$\small v_{sufficient}=\sqrt{4gL}(\frac{m+M}{m})$