IIT JEE , AIEEE Forum - Articles Category , Community shelf

mirtunjay_svm2007

We can apply the rules for connecting translational motion to rotational motion by remembering that s = r*theta. This connection works for any solid object rotating around a fixed axis. If a rolling object of unit radius (r = 1) rotates through one revolution, the arc-length traversed is 2*Pi and the horizontal distance traveled is 2*Pi. Try viewing it through the eyes of Maple.

We can take as an example of the use of such esoteric knowledge, a mystery in Egyptology. The pyramids of ancient Egypt are built with a square base and equal length sides. A measurement of the angle of inclination of the sides w.r.t. the horizontal shows them to be at an angle of 51° 51'. All but one of the large pyramids has this as the angle of inclination. The strange value of this angle results from the design of the shape of the pyramid such that the ratio of the circumference of the base to the height of the apex is equal to 2*Pi. Since the Egyptians probably did not know about Pi, it was a mystery as to how they could design a structure which includes it so precisely.

One possible model for how this resulted comes from T.E. Connely. The proposal is that the sides of the pyramid were laid out by rolling a drum of diameter L an integral number of times along the ground, turning the drum by 90° , then rolling the same number of revolutions, etc. until the square base is laid out. Then, rolling the drum by half the number of revolutions for a side, turning by 90° and rolling again for half the number of revolutions puts the drum at the center of the square. Building the pyramid so that the maximum height is an integral number of drum diameters high ensures that the circumference of the base compared to the height is a ratio of integers as shown in the figure below.

If we say that the number of turns used to form one side of the square base is 2n, then the circumference is 4*(2n)*Pi*L. If the height is built so that it is 4nL high, then the ratio of circumference to height is 8*n*Pi*L/(4nL) = 2*Pi! No knowledge of the value of Pi is necessary.

Examples of Rotational Dynamics

We can begin with two problems, one familiar, another which is unfamiliar and extremely difficult to do correctly without making use of our notions of rotational motion.

A block of mass m₂ is attached by a massless, stretchless string to a mass m₁. The string goes over a pulley of mass m₃ and radius R. The pulley has the shape of a flat disk. Find the accelerations of the masses and the tensions, T₁ and T₂.

Solution: First, note that the tensions must have different magnitudes since the string acts on all three objects. From the diagram of forces acting on the pulley, we can see that T₁ should not equal T₂ since the pulley definitely turns with the string as mass m₁ goes down and mass m₂ goes to the right. Next, we should note that the accelerations of the two masses are the same as long the string does not change length. Finally, we note that the conventions for positive directions as shown on the free-body diagrams leads us to conclude that the signs of the accelerations for m₁ and m₂ will be the same, but the angular acceleration of the pulley will be opposite sign since positive rotation of the pulley would correspond to negative acceleration of the masses. If the string does not slip on the pulley surface, then the angular acceleration is alpha = -a/R where a is the acceleration of the masses. The free-body diagrams and the calculations of the torques yield

1: m₁g - T₁ = m₁a

2: T₂ = m₂a

3: T₂R - T₁R = I₃alpha

= ½ m₃R²(-a/R)

T₁ - T₂ = ½ m₃a

The simplest approach to a solution is to just add all 3 equations together. The tensions cancel out to yield
m₁g = (m₁ + m₂ + ½ m₃)a ==>
a = m₁g/(m₁ + m₂ + ½ m₃)
The tensions are then

T₁ = m₁(g - a)

= m₁g[( m₂ + ½ m₃)/(m₁ + m₂ + ½ m₃)]

T₂ = m₂ m₁g/( m₁ + m₂ + ½ m₃)
A thin cylindrical rod of mass M and length L is free to pivot around one end. The rod is initially resting with its center-of-mass directly above the pivot point as shown in the figure below.

If the rod is given a very slight push to the right, what is the velocity of the center-of-mass of the rod when its center-of-mass is at the level of the pivot, i.e. the rod is horizontal?
Solution: We start by noting that the motion of the rod is purely rotational. The center-of-mass does not translate, but simply rotates around the fixed axis of rotation at the pivot, maintaining a constant distance of L/2 away from the axis. We can apply the work-energy theorem most easily here to get the speed. The work is done by gravity acting on the center-of-mass. Since we are allowed to calculate the work done by any means convenient to us, we choose to use the translational form of the work done. The center-of-mass moves from a height of L/2 above the pivot to the level of the pivot, so the work done by gravity is Mg(L/2). Therefore, noting that the rotational inertia of a thin cylinder rotating around its end is (1/3)ML² and that the rotational velocity of the center-of-mass around the axis of rotation is omega = v/r = v/(L/2), we find

W = K_f - K₀

Mg(L/2) = ½ I*omega² - 0

= ½ (1/3)ML²*(v/(L/2))² ==>

Mg(L/2) = (2/3)Mv² ==>

v = sqrt[(3/4)gL]

Community Contributions - Articles by goIITians