Stuck, all I have is the first part which is equation (8.13), L=1/2MR'^2+(1/2mue(r'^2)-U(r))

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The moment of inertia for a solid sphere is given by I = 2/5MR squared and a solid disk is given by I = 1/2MR squared. If both objects have the same mass and radius and both are released from rest at the same point in an inclined plane, which would reach the bottom first?

A particle of mass m moves under the influence of a central force.The potential energy function is
V(r) = {
K (a2âr2) /2 if r â¤ a
0 if r > a

as illustrated in the figure. The initial position is (x,y) = (a/2,0) and the initial velocity is (vx,vy) = (0, v0).
[Data: m = 1.0 kg; K =17.1 N/m; a = 1.0 m; r0 = 0.5 a; v0 = 2.0m/s.]

(A) Determine the time ta when the mass crosses the boundary r =a.

(B) Determine angular position Ïa where the particle crosses theboundary r = a.

***HINT GIVEN***

Use the constants of the motion to analyze the orbit.

Write L = mr^2 dÏ/dt = the initial value.
What is the initial value?

Write E = (1/2)m (dr/dt)^2 + L^2/(2mr2) +V(r )
= the initial value. What is the
initial value?

Part A. In the energy equation,
change the variable from r to u,
where r =sqrt(u). The result can be
solved by u(t) = c1+c2 cosh(Î±t).
Determine c1, c2, and Î±.

Part B.
Ï = Integral 0 to t [L/(mr^2)] dt

Suppose there is a little block attached to a cord passing overapulley through a hole in the

show all calculations and work ...THANKS!

Suppose there is a little block attached to a cord passingovera pulley through a hole in the horizontal frictionlesssurface. Atfirst the little block is revolving with an initialangular speed of6rad/sec at a distance r from the center of thathole. So if thischord is pulled from below until the new radius isr/6, then whatwill be the new angular velocity?

Show all calculations and work ...THANKS!

w=work andw(i)=initial work

r= radius

m=mass

Hints given: L= r*p- = r*m *v(1)

v- =w*r

L(i)=r(i)*m*w(i)*r(i)

Eventually masses will cancel out so leftwithr(i)^2*w(i)--> solve for w(i)