Frequency of small oscillation of ball hanging in accelerating car

Problem:

A car accelerating at uniform rate a_o has a small ball hung under the roof by a string of length L. What is the angle \theta between the string and the vertical direction when the ball is is steady state? What will be the frequency of oscillation if the ball is set in small oscillation?

Solution:

Steps:

  1.  Specify the chosen reference frame
  2. Sketch the free body diagram
  3. Sum the forces along x and along y
  4. Solve for theta

 

– Frequency of small oscillation of ball hanging in accelerating car

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Amount of water that would make the system most stable

Problem:

An empty cylindrical container has height L, mass m and inner radius r. The center of gravity of the container is located at height h from its bottom. How much water (in terms of height) should one pour into the container to make it most stable? The density of water is \rho.

Solution:

The mass of water adds up to the stability of the system but too much water would make the center of mass high (in terms of height). You need basic knowledge on:

  1. Solving differential calculus – quotient rule
  2. Solving quadratic equation
  3. Finding center of mass

Steps:

  1. Find the center of mass of the system (cylinder + water)
  2. Minimize the center of mass in terms of height of water
  3. Solve for height of water

Try to solve the problem first. If you’re hopeless, then here it is

pset1-3 – height of water with most stable system

 

Rod hanging by a string next to wall

Problem:

A uniform rod AB with length l and weight W is placed as shown in the figure, with end A touching a rough wall and end B hanging by a strong, inextensible string. The other end of the string is fixed at point C on the wall is \mu. The rod is hung horizontally and the angle between the string and the rod is \theta. Derive the condition that \mu and \theta should satisfy such that the rod stays at equilibrium.

Solution:

Steps:

  1. Sketch the free body diagram
  2. Sum forces along x
  3. Sum forces along y
  4. Sum torque
  5. Solve system of equations for \theta and \mu

Try to solve the problem first without looking at the solution, if you’re still helpless,

Here it is:

pset1-2 – Rod hanging by a string

Acceleration of Missile

Problem:
Imagine a tank traveling with constant speed v_1 along the direction AB, and a missile is chasing the tank with constant speed v_2. The direction of motion of the missile is always pointing towards the tank. At time t the tank is at position F and the missile is at position D, where FD is perpendicular to AB and FD=L. What is the acceleration of the missile at this instant?

There’s a lot of solutions to this problem. But I will provide the formal one (long method). This method requires knowledge skills on:

  1. Differential Calculus
  2. Vector Algebra
  3. Distance-Velocity-Acceleration knowledge

Steps would include:

  1. Writing initial conditions for missile and tank
  2. Write the velocity of tank as magnitude*direction
  3. Differentiate the velocity to get acceleration of missile
  4. Substitute the initial conditions
  5. Simplify
  6. Answer should be a=\frac{v_1 v_2 }{L} \hat{x}

Try to solve the problem first without looking at the solution. If you’re still helpless, then here it is:

pset1-1 – acceleration of missile

Derivation of Kinetic energy differential equations

PROBLEM:
1.1 Goldstein Classical Mechanics 3rd ed.
Show that for a single particle with constant mass the equation of motion
implies the following differential equation for the kinetic energy:

\displaystyle \frac{dT}{dt} = \bf F \cdot \bf v

while if the mass varies with time the corresponding equation is

\displaystyle \frac{d(mT)}{dt} = \bf F \cdot \bf p

SOLUTION:
The kinetic energy T is given by

\displaystyle T = \frac{1}{2}m\bf v^2

The time derivative of kinetic energy is

\displaystyle \frac{dT}{dt} = \frac{d}{dt} (\frac{1}{2}m\bf v^2)

chain rule

\displaystyle \frac{dT}{dt} = m\mathbf{v}\cdot\frac{d\bf v}{dt}

commutative property of dot product

\displaystyle \frac{dT}{dt} = m\frac{d\bf v}{dt} \cdot \bf v

\displaystyle \frac{dT}{dt} = m\bf a \cdot \bf v

Newton’s second law

\displaystyle \frac{dT}{dt} = \bf F \cdot \bf v

When the mass varies with time, again the kinetic energy T is given by

\displaystyle T = \frac{1}{2}m\bf v^2

Multiply both sides by m

\displaystyle mT = \frac{1}{2}m^2\bf v^2

\displaystyle mT = \frac{1}{2}{(m\bf v)}^2

\displaystyle mT = \frac{1}{2}\bf p^2

The time derivative is

\displaystyle \frac{d(mT)}{dt} = \frac{d}{dt} \frac{\mathbf p^2}{2}

\displaystyle \frac{d(mT)}{dt} = \mathbf{p} \cdot \frac{d\bf p}{dt}

\displaystyle \frac{d(mT)}{dt} = \mathbf{p} \cdot \bf F

\displaystyle \frac{d(mT)}{dt} = \bf F \cdot \bf p

Comment for corrections/suggestions and further questions. Thanks!