The coservation of energy principle

[AvrGang]

Homework Statement

I had just finished solving 9 energy problems and I'm stuck with this hard question:

A simple pendulum, which consisits of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.00 meters long and makes an initial angle of 30.0 (degrees) with the vertical, calculate the speed of the particle(a) at the lowest point in its trajectory (b) when the angle is 15.0(degrees).

I didn't understand the whole idea of this question ( I mean I can't draw it in my mind)

so please i'll appreciate any help

thnx

Homework Equations

The law of conservation of energy:
M.E.(f) = M.E.(i)

K.E.(f) + P.E.(f) = K.E.(i) + P.E.(i)

K.E.( kinetic energy) = 1/2 m(V)^2
P.E.(gravetational potential energy) = mgh or (P.E.=mgy)

I think there is no elastic potential energey

The Attempt at a Solution

I didn't understand the idea of question

 

[AvrGang]

I'm online for any discussion

 

[Sourabh N]

Choose a reference line to calculate PE, and equate total energy initially and finally(at lowest point in first case).

 

[AvrGang]

ok but what is the initial height

How can i calculate the height

 

[Sourabh N]

height - distance of the point from reference line.

 

[AvrGang]

I did that: my reference is the lowest point of the trajectory

K.E.(f) + P.E.(f) = K.E.(i) + P.E.(i)

(1/2 m(V)^2 + mgh)(f) = (1/2 m(V)^2 + mgh)(i)

1/2 m(V(f))^2 = mgh(i)

how can i find initial hieght h(i)

 

[AvrGang]

can we use the string's length?

 

[Sourabh N]

Nope. Use trigonometry. Initially string is at 30 deg with vertical, can you calculate height using this(remember, height is of bob)?

 

[AvrGang]

ok is this the answer of a)

K.E.(f) + P.E.(f) = K.E.(i) + P.E.(i)

(1/2 m(V)^2 + mgh)(f) = (1/2 m(V)^2 + mgh)(i)

1/2 m(V(f))^2 = mgh(i)

1/2 m(V(f))^2 = mg(5sin30)

(V(f))^2 = 2g2.5

V = 7m/s

 

[AvrGang]

right or wrong?

please tell me

I'll be very thankful

 

[Sourabh N]

Well, how did you get h(i) = 5 sin30 (remember your ref line is at bottom of trajectory, so h(i) is just the vertical distance of bob at initial time from ref line)

 

[AvrGang]

I used trigonemotry to find it as you say

 

[AvrGang]

try it i think my answer may be correct

 

[Sourabh N]

h(i) = l (1 - cos $$\theta$$$$_{i}$$)

 

[Sourabh N]

Convince yourself about this or ask.

 

[AvrGang]

ok now I can solve the question using this

but i still don't know from where did you bring (1-cos$$\theta$$)