Pendulum Problem?

Question

I just don't get these pendulum problems. Could someone tell me how to go about them...

A pendulum is formed by attaching a mass, m = 0.25 kg to the end of a string of length L = 1 m. The other end of the string is attached to a nail so the pendulum can swing in a complete circle in the vertical plane. There is no friction between the nail and the string.

The mass is released from rest with the string at a 60° angle with respect to the vertical as shown. What is the speed, v, of the mass when the angle between the string and the vertical is 30°?

a)v = 1.52 m/s
b)v = 1.69 m/s
c)v = 1.81 m/s
d)v = 2.12 m/s
e)v = 2.68 m/s

The pendulum is now set in motion in such a way that it swings in a complete vertical circle with the tension of the string being equal to the weight of the mass at the top of the circle. What is square of the speed, vb2, of the mass at the bottom of the swing (i.e., the lowest point in the swing)?

a)vb2 = 3gL/2
b)vb2 = 2gL
c)vb2 = 5gL/2
d)vb2 = 4gL
e)vb2 = 6gL

Answer 1

the loss of PE is the gain in KE

m*g*h=.5*m*v^2

v=sqrt(2*g*h)



for this we need h

h=L*((1-cos60)-(1-cos30))

h=L*(cos30-cos60)

e) 2.68



at the top

vt^2/L=g

vt^2=L*g

the starting KE less gain in PE is the KE at the top



.5*m*vb^2-m*g*2*L=.5*m*vt^2



simplify

vb^2=5*L*g

I don't see the answer, but I believe this is correct.



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Answer 2

Potential energy = Kenetic energy



PE = KE



where PE is (1/2) (mass) (velocity sqrd) = 1/2mV^2

and KE is (mass) (gravity constant) (height) = mgh



giving:



1/2mV^2 = mgh



mass is constant... canceling out



1/2V^2 = gh



where h is the height of the ball at 30 degrees

and h is the sin function or y coordinate



giving 1/2 V^2= g (sin 60 -sin30)



giving 1/2 V^2 = 9.8(.366)



giving 1/2 V^2 =3.59



multiply both sides of equation by 2



giving V^2 = 7.18



giving V = sqrt 7.18 = 2.68m/s which is answer E



part two: velocity at bottom of swing if starting from verticle



Where Vb is velocity at bottom of swing

V is used in the calculation for Vb



1/2 mV^2 = mgh



m is constant and cancels out



giving 1/2V^2 = gh



where h is the length of the string at the bottom



h = L = 1meter or 1 in the calculation or L for the purpuses of the answers



giving 1/2 V^2 = gL



giving V^2 = 2gL this being answer number B (Vb^2 = 2gL)

Answer 3

For small masses, the formula for the period of a pendulum is: T = 2* pi * sqrt (L/g) Divide the length of the rope by the gravitational acceleration (9.8 m/sec^2), take the square root, and multiply by 2 pi. Except in this case, you already know the period and don't know the length. Rearrange the equation to solve for L T/(2 pi) = sqrt (L/g) T^2/(4 pi^2) = L/g gT^2/(4 pi^2) = L