Help me with this physics question involving work?

Question

A ball is attached to a horizontal cord of length L whose other end is fixed
(a) If the ball is released, what will be its speed at the lowest point of its path? (Answer using g for acceleration due to gravity and L as necessary.)
(b) A peg is located a distance h directly below the point of attachment of the cord. If h = 0.86 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?

Answer 1

a) Since the energy gained from descending in a gravity field is constant regardless of whether the object is in free fall or is a weight on a pendulum, we can use the standard 'SUVAT'; equations for this one:
s = displacement = L, (since this is the distance it will have descended)
u = initial velocity = 0
v = final velocity
a = acceleration due to gravity = g

v^2 = u^2 + 2as

v^2 = 0 + 2 x g x L
So, v (its speed) = (2gL)^1/2

b) If h is 0.86L below L, then the length of cord below h is (1-0.86)L = 0.14L.
So, 0.14L is now rotating the weight around peg 'h'.
Use the same equation again, but this time, because the pendulum is rising against gravity, use a = '-g'

S = 2 x 0.14L = 0.28L
U = The velocity the pendulum gained in part a) = (2gL)^1/2
V = V
A = -g

V^2 = U^2 + 2AS
By substituting in the above values:
V^2 = ((2gL)^1/2)^2 -2g x 0.28L
Simplify:
V^2 = 2(gL - 0.28gL)
V = (2(gL - 0.28gL))^1/2 = (2(0.72gL)^1/2 = (1.44gL)^1/2

Answer 2

a.) This first problem deals with conservation of energy. At the top, the ball-cord system has a total potential energy of mgh, where h is the maximum amount of distance that the ball can fall which in this case is L since it is the length of the cord that inhibits the ball from falling further. At the lowest point, all the potential energy is converted to kinetic energy so. E=mgh=1/2mv^2

mgh=1/2mv^2
gh=1/2v^2
2gL=v^2
sqr(2gL)=v

b.) This problem also deals with conservation of energy. This time we are going from potential to potential+kinetic. At the top we start with mgL potential. At the bottom we will have 1/2mv^2 potential. At the top of the circular path about the peg with will have 2mgh'+1/2mv^2 energy. (h'=(L-h) height since the peg is (L-h) above the floor and 2 because it's the diameter the ball is travelling which is twice the radius; the radius being the height above the floor the peg is at. This term has to be added since the peg inhibits the ball from reaching it's inital height of h=L.) The sum of all the energy at any instant is the total energy. The total enegy of this system is the same as the inital total energy which is mgL. So, we say: E=mgL=2mgh'+1/2mv^2

mgL=2mgh'+1/2mv^2
gL=2gh'+1/2v^2
gL=2g(L-0.86 L)+1/2v^2
gL-2g(.14 L)=1/2v^2
gL-.28gL=1/2v^2
.72gL=1/2v^2
1.44gL=v^2
sqr(1.44gL)=v
1.2sqr(gL)=v

Answer 3

Use Energy to solve this problem

Conservative Energy:

Etotal = Ekinetic + Epotential

In this problem the motion of the ball is the transfer energy between kinetic and potential. When the ball released at horizontally, there is no kinetic at that point. Only potential energy which is mgh or mgL. When the ball moving down, potential energy transfers to kinetic energy which is 1/2mv^2

a) the speed is 1/2mv^2 = mgL
v = square root of (2gL)
b) At the moment when it reach the top of its circular path, kinetic gone and only potential left. Thus v = 0 at that point. You hard to see it because it happens just slit of second.

Answer 4

Ignoring work done by non-conservative forces....

Ef = Ei

KEf + PEf = KEi + PEi

KEf + 0 = 0 + PEi

½m v² = mgh

Since the ball starts horizontally, it's initial height is L.

the m's cancel, and we get

½ v² = gL

v = sqrt(2 gL)

For part 2 of your question, we have

KEf + PEf = KEi + PEi

½ mv² + mg(L - 0.86 L ) = 0 + mgL

½ v² + g(0.14 L) = gL
½ v² + 0.14 gL = gL
½ v² = 0.86 gL
v² = 1.72 gL

v = sqrt(1.72 gL) = 4.11 sqrt(L)