Fundamental Physics by Ramamurti Shankar at Yale_4_Rotations

This chapter is about rotations. Omega – angular velocity is introduced to study this topic. Based on angular velocity, radian is also used to get a quick measurement of the constant value or unit to calculate the circular distance.

Professor Shankar list linear and circular motion side by side to help understand the concepts. What’s worth noting is that for a blade point on this rotating saw, there are two acceleration, one is the normal circular motion acceleration, that causes the direction of tangible speed change; the other is the tangible acceleration, pertinent to the time when the blade is speeding up or slowing down.

Similarly we deduce the kinetic energy for a circular moving rigid body and define I as the momentum of inertia. Comparable to the linear motion, kinetic energy = 1/2*mV^2. I is not only related to the summation of m, but also the radius R.

When exert an external force(perpendicular) to the rigid body, causing it to move in circular motion, we can apply the change of kinetic energy is equal to the work done(delta w), consequently, we get F*radian, which is given a name Torque. Toruqe can be summed up if there are multiple masses.

The complete formula would be as following since we also consider the common cases are that the F is not perpendicular to the surface, but with an angle, while the vector, cos part is of the same direction to the center, it doesn’t yield any effective output for moving any angles or radians. As analogues to linear motion, work done/delta w = F*deltax, power of this work p = F*v; work done in circular motion deltaw = Torque*delta theta (is deduced from F*radian*delta theta, remember?.), power or efficiency of this delta work p = Torque*(delta theta/delta t) = Torque*alpha.

In the following it a real and simple problem we can apply this knowledge of Torque.

Next, let’s tackle with the problem of calculating the momentum of inertia of any irregular shaped piece of rigid object, using integral, it’s straightforward to deduce I at the center of the object(surface) as well as at any location of it. However, there is a shorthand for calculating this metric, which can be defined as the inertia momentum of the center point + mass of the whole piece * distance to the center of mass squared. This is called Parallel Axis Theorem.

How to prove it is correct?

First depict a new pivoting point, a randomly chosen point as the infinitesimal point to conduct integral computation on the surface and draw the relation between these two points and the center point.

Then we apply the formula to calculate the inertia as usual. Unfolding the bracket, we get the below parts. Very critical here, the middle product = 2 d vector * sum up of mi*r vector vanished because the summation of mi * r vector is zero, it’s origin point, the r vector cancel each other out in terms directions.

Applying this to study the moving of a car tire, kinetic energy = 1/2(MR^2 +1/2MR^2)*W^2, which is also 1/2(I road) W^2 ( the point touching the ground’s inertia). If we take a deeper peek, at every second, the top point of the tire is forging on with 2V while the bottom point is still.

Now to study the problem of figuring out the speed of a ball at a height of h on a slope, down to the bottom, it’s not simply mgh = (1/2)mV^2 anymore, the ball should be treated not as a point but a rigid object, so what applicable is

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