Acceleration Due to Gravity . Moment of Inertia of a Rigid Body

Acceleration due to gravity . Moment of inertia of a rigid body

EXERCISE 1

Acceleration due to gravity

1.1. Introduction

A simple pendulum consists of a small object (the pendulum bob) suspended from the end of a lightweight cord. The cord doesn’t stretch and the mass of the bob must be small relative to the length of the cord. The motion of a pendulum moving back and forth with negligible friction resembles simple harmonic motion. The pendulum oscillates along the arc of a circle with equal amplitude on either side of its equilibrium point (where it hangs vertically) and as it passes through the equilibrium point it has its maximum speed.

When the cord makes an angle  to the vertical, then we can resolve the acceleration g of the bob situated in gravitational field into two components: one along the cord is N, and the second one is tangential to the trajectory S (S=-g sin). The bob is considered to be a material point.

Fig.1.1 Simple pendulum

The displacement of the pendulum along the arc is given by:

s = l.  (1.1)

where  is the angle the cord makes with the vertical and l is the length of the cord, as shown on Fig.1.1.

Tangential acceleration is:

(1.2)

for small angles we can assume:

 and s=x (1.3)

From (1.2) we have:

or i.e. an equation of SHM(1.4)

We observe simple harmonic motion of the pendulum for small angles . By taking into account that and we can write equation for the period of a simple pendulum as:

(1.5)

where: l – is the length of the pendulum

g – is the acceleration due to gravity

Finally we get:

(1.6)

1.2. Measurements

Take a series of measurements of the period T1 at any length of pendulum. Then shorten or lengthen the length of pendulum about well-known value D, and measure new period T2.

Since:

(1.7)

and

(1.8)

so

(1.9)

(1.10)

We can make such an approximation because the oscillation of the pendulum is the damped oscillation. Amplitude decreases with time to zero. The period is independent of amplitude and doesn’t change with time. The differences between T are equal even though the amplitude fades. The formula (1.5) holds only for small amplitudes /<5 o /.

To get T, you have to measure /5-10 times/ time of several /50-100/ periods and calculate average value to find T.

The data should be collected in a Table 1.1:

Table 1.1.

1.3. Results, calculations and uncertainty.

From the data collected in the table calculate the average values of

(1.11)

and

(1.12)

Estimate the uncertainty of the measured values by calculating the total differential of equation (1.10) respectively:

(1.13)

The final result reads:

(1.14)

1.4. Questions

1. What kind of assumptions one has to make to arrive at SHM equation?
2. Explain the “source” of approximation . Show the condition under which we can accept it!
3. Can you predict the value of oscillation period of the Mars?
4. What about the value of T in case when the pendulum moves vertically upward with acceleration “a”?
5. What is a physical pendulum? Derive an equation for its period!

1.5 References

1. Szczeniowski S., Fizyka Doświadczalna, Część I, Mechanika i Akustyka, PWN, Warszawa, 1980
2. Orear J., Fizyka, Tom I, WNT, Warszawa 1990
3. Resnick R., Halliday D., Fizyka, Tom I, PWN, Warszawa,1966
4. Szydłowski H., Pracownia fizyczna, PWN, Warszawa, 1994
5. GiancoliD.C., Physics. Principles with Applications, Prentice Hall, 2000
6. Young H.D., Freedman R.A., University Physics with Modern Physics, Addison-Wesley Publishing Company, 2000

EXERCISE 2

Moment of inertia of a rigid body

2.1. Introduction

Let us consider a rotating rigid body, such as a wheel rotating about an axis through its center, such as an axle. We can think of the wheel as consisting of many particles located at various distances from the axis of rotation R1,R2,…..Rn.

The moment of inertia (or rotational inertia) of the body is represented by equation:

[kg m2] (2.1)

The sum represents the sum of the masses of each particle in the body multiplied by the square of the distance of that particle from the axis of rotation. As can be seen from eq.(2.1) the rotational inertia of an object depends not only on its mass, but also on how that mass is distributed with respect to the axis. For example a large-diameter cylinder will have greater rotational inertia than one of equal mass but smaller diameter. The moment of inertia of a given system is different also for different axes of rotation.

Many bodies can be considered as object with a continuous distribution of mass. In this case eq.(2.1) defining moment of inertia becomes:

(2.2)

where represents the mass of any infinitesimal particle of the body and R is the perpendicular distance of this particle from the axis of rotation. The integral is taken over the whole body (usually it is double or triple integral).

2.2. Measurements

Put the body of mass on the disc. The radius of the disc is . The disc is hanging on three vertical strings. The strings have the same length l and they are fastened to the disc at the same distance R from the center O of the disc and they form the equilateral triangle. The central axel of inertia of the body should superimpose the axle OO” of the disc like on Fig.2.1

Fig.2.1

Now rotate the disc by a small angle  so the point C is now at the place of the point D, the center of gravity of the hanging masses is slightly raising by value z. If you release the disc, it starts vibrate with the period:

(2.3)

where: -mass of the disc,

-momentum of inertia of the disc,

Since you know R and l you can calculate the constant , and then from the formula (see Table 3):

(2.4)

After taking measurements of T, you can calculate from the formula (2.3).

Find the moment of inertia of the wooden block with respect to its center axle of inertia.

Compare it with the moment of inertia calculated from the formula, which takes into account the sizes of the edges and mass of the block.

Fig.2.2 Rectangular thin plate

Moment of inertia for rectangular thin plate of uniform composition (like in Fig.2.2), of length L, and width W is:

(2.5)

By analogy find the moment of inertia for the rectangular steel block and wooden disc like on Fig.2.3.

Fig.2.3 Solid cylinder

The moment of inertia of the solid cylinder with radius r is:

(2.6)

Take the measurements of T six times for each rigid body. The data should be collected in a table.

2.3. Results, calculations and uncertainty

From the data collected in the table 2.1 calculate the average value of

Table 2.1.

(2.7)

Estimate the uncertainty of moments of inertia by calculating the total differential of equation (2.3)

(2.8)

The final result reads:

(2.9)

Compare moment of inertia calculated from experimental data and moment of inertia calculated from formula, which take into account the sizes and mass of examinated bodies.

2.4. Questions

1. Show that the moment of inertia of a uniform hollow cylinder of inner radius R1, outer radius R2 and mass M, is equal to I = M(R12+R22) if the rotation axis is through the center along the axis of symmetry.
2. Comment on angular momentum and its conservation.
3. Derive and comment on rotational kinetic energy.
4. What is the torque good for?
5. Derive the formula for work and power for the body rotating about a fixed axis.

2.5. References

1. Szczeniowski S., Fizyka Doświadczalna, Część I, Mechanika i Akustyka, PWN, Warszawa, 1980
2. Orear J., Fizyka, Tom I, WNT, Warszawa 1990
3. Resnick R., Halliday D., Fizyka, Tom I, PWN, Warszawa,1966
4. Szydłowski H., Pracownia fizyczna, PWN, Warszawa, 1994
5. GiancoliD.C., Physics. Principles with Applications, Prentice Hall, 2000
6. Young H.D., Freedman R.A., University Physics with Modern Physics, Addison-Wesley Publishing Company, 2000

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