Solutions for Problem Set 1

Last revised 1999/09/09

HRW pp. 11-13 #1, 16, 39
Sheet # I and II

Sheet #I:

    (a)

Quantities given and needed:

Period T = ?
Length L
Acceleration of gravity g

Technique:

We can get the required relation, up to an overall constant, by units. Try the general form

T = K La gb , with K unitless

Details:

Using the units of the equation for T we get

[sec] = [m]a [ m/sec2 ]b

or

[sec]1 [m]0 = [m]a + b [sec]-2b

Since both of the units must work out correctly:
1 = -2b 0 = a + b

Hence b = - 1/2, a = 1/2, and

T = K (L/g)1/2

    (b) [Your data]

    (c) You should get T = 2 pi (L/g)1/2

Sheet #II.

Quantities given and needed:

Height H [m]
(Horizontal) distance of flight R = ? [m]
Initial speed v [m/sec]
Acceleration of gravity g [m/sec2 ]

Technique:

As usual, try the general form:

R = K Ha vb gc

Details:

The units of the general form work out to be

[m] = [m]a [m/sec]b [m/sec2 ]c

so that

[m] = [m]a+b+c [sec]-b-2c

Since both m and sec must agree

1 = a + b + c

and

0 = -b - 2c

We can solve for b and substitute into the first equation to get

b = -2c
a - c =  1

Since we would need three equations to solve for three unknowns, units alone do not solve the problem! We can reduce the number of unkowns to one, and we will choose a and b to eliminate. We can also restrict the range of c by finding which exponents have to be positive and which negative:

  1. If we want to throw a rock a long way, we certainly want a large initial velocity, so b>0.
  2. Being on the top of a large tower would help, so a>0 .
  3. A large gravitational force would certainly hurt, so c<0 , which is consistent with b>0

Then a>0 applied to c = a - 1 means c>-1. All told, the best we can do towards solving the equation is

R = K H1 - |c| v2|c| / g|c|

0 < |c| < 1

Only by guessing that c might be half-integral can we get to the form

R = K v (H/g) 1/2

Even though we cannot derive this formula by units, we can use the units argument to help remember the result.