Question

Help Needed! (See attachment)

Answer 1

determine the condition that must be satisfied for this to take place.

As the ball swings around the peg the speed must be great enough to offset the force of gravity as the ball reaches its maximum height. i.e. the centripital force must equal the weight.

can you take it from here?

Answer 2

No, sorry.

Answer 3

Ok,

centripetal force = weight of the ball
\[mv ^{2}/r = mg\] therefore
\[v=\sqrt{gr}\] where v is the velocity of the ball as the string strikes the peg and \[r=11-d\] therefore\[v=\sqrt{g \left( 11-d \right)}\]
this is the condition that must be satisfied.

Now use the conservation of energy to determine the relation ship between the height of the ball above its lowest position, angle of the string and mass to the velocity at the bottom as the string strikes the peg.\[mgh=\frac{ 1 }{ 2 }mv ^{2}\]and\[h=11\cos(49\deg)\] Now substitute our velocity condition and the value of h into the conservation of energy equation and solve for d.

Answer 4

I got d = -3.433...

Answer 5

Sorry, I made mistakes in the equations and approach that I suggested above. first\[h=11-11\cos(49)\]second the velocity as the peg is struck is not the same velocity at the top of the trajectory. at the top of the trajectory we have both potential and kinetic energy so the total energy at this point is\[E( apex )=mg2r+mV ^{2}/2 = E(start)=mgh =mv ^{2}/2\]the mass must have enough momentum to carry it over the top. \[V <v\] Note that the answer will be independent of the mass if my approach is correct. Hope I'm right this time.

Answer 6

Almost forgot remember at the top (apex) of the trajectory\[mV ^{2}/r=mg\] thew centripetal force = weight of the mass so the mass coasts over the top keeping the radius of its path constant.