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20 Nov 2019

For this problem. I understand how to set it up and how to conserve the mechanical energy so we will have two sides an equation. But when I try to get the potential energy at the final state, where the string forms an angle Θ, the equation you have to use is mgL(1-cosΘ) instead of just mgL(cosΘ). So how do you figure out to use the equation mgL(1-cosΘ)?

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Hubert Koch
Hubert KochLv2
6 Aug 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

Three blocks of weights W1 = 11 lb, W2 = 9 lb, and W3 = 10 lbare strung in the massless pulley system shown and released fromrest at initial time t0 at a distance H = 3 ft above the stopper asshown. The stopper has a hole in it which allows the narrower block3 to pass through. However, when block 2 contacts the stopper,block 2's motion stops, and only blocks 1 and 3 continue tomove.

a) What is the change in gravitational potential energy of thesystem from the initial time t0 to the time t1 immediately beforeblock 2 contacts the stopper?

HELP: Use kinematics to relate the positions and velocities of theblocks.

b)What is the kinetic energy of the system at time t1 immediatelybefore it contacts the stopper?

HELP: What is the initial kinetic energy of the system?
HELP: Conserve energy from time t0 to time t1 just before block 2contacts the stopper.

c)What is the speed of block 2 at time t1 immediately before itcontacts the stopper?

HELP: This problem involves translation only, so you may usekinetic energy T=(mv^2) /2 for each block.

d)What is the kinetic energy of the system at time t2 just afterblock 2 contacts the stopper, i.e. the system in which blocks 1 and3 continue their motion without block 2?

HELP: Is (mechanical) energy conserved during the collision betweenblock 2 and the stopper from time t1 to time t2?
HELP: No, it is not. Has the moving mass of the systemchanged?
HELP: Have the speeds of the blocks that continue to movechanged?

e)When block 3 first comes to rest at time t3 (after passingthrough the stopper), how far will it be from where itstarted?

HELP: Conserve energy from the time immediately after the collisionbetween block 2 and the stopper to the time when blocks 1 and 3stop, i.e. from time t2 to time t3.
HELP: Carefully consider for what the question is asking.


Learning Goal: Understand how to determine the constants in thegeneral equation for simple harmonic motion, in terms of giveninitial conditions.
A common problem in physics is to match the particular initialconditions - generally given as an initial position x_0 andvelocity v_) at t=0 - once you have obtained the general solution.You have dealt with this problem in kinematics, where theformula

1. x(t) = x_0 + v_0t + at^2

has two arbitrary constants (technically constants of integrationthat arise when finding the position given that the acceleration isa constant). The constants in this case are the initial positionand velocity, so "fitting" the general solution to the initialconditions is very simple.

For simple harmonic motion, it is more difficult to fit the initialconditions, which we take to be

x_0, the position of the oscillator at t = 0, and
v_0, the velocity of the oscillator at t = 0.

There are two common forms for the general solution for theposition of a harmonic oscillator as a function of time t:

2. x(t) = A\cos(omega t + phi) and
3. x(t) = C\cos(omega t) + S\sin(omega t),

where A, phi, C, and S are constants, omega is the oscillationfrequency, and t is time.

Although both expressions have two arbitrary constants--parametersthat can be adjusted to fit the solution to the initialconditions--Equation 3 is much easier to use to accommodate x_0 andv_0. (Equation 2 would be appropriate if the initial conditionswere specified as the total energy and the time of the first zerocrossing, for example.)

Find C and S in terms of the initial position and velocity of theoscillator.
Give your answers in terms of x_0, v_0, and omega. Separate youranswers with a comma.

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