PHYS 1003 Lecture Notes - Lecture 15: Cross Product, International System Of Units, Rotational Energy

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scarletmouse10

10 Jan 2017

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Document Summary

Cross product of the distance from the center of rotation and the force applied at a point. Work done by a constant torque is angular displacement. Integrate the torque function with respect to the angle. Work done on a system is equal to the rotational kinetic energy. Rotational inertia for a rotating body r is always measured relative to the center of rotation. Rotational inertia for a continuous mass distribution we know the shape, the density but not the coordinates of every individual particle. If we integrate the mass distribution over a volume. The mass f each cylinder is where dr is the thickness. The moment of inertia of each cylinder is. To find the total moment of inertia. If several geometries are rotating around the same axis, then the combined rotational inertia is the sum of the individual rotational inertias.

Related Questions

Consider a turntable to be a circular disk of the moment of inertia I_t rotating at a constant angular velocity ωi(note that angular velocities use the Greek letter omega and not double-u) around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry") as shown in(Figure 1). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.

Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is Ir. The initial angular velocity of the second disk is zero.

There is friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular velocity.

Part A

What is the final angular velocity, ωf, of the two disks?

Express ωf (omega subscript f) in terms of It, Ir, and ωi (omega subscript i).

Part B

Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, Kf, of the two spinning disks?

taupehamster183

The moment of inertia of a uniform-density disk rotating about anaxle through its center can be shown to be (1/2)MR2. This result isobtained by using integral calculus to add up the contributions ofall the atoms in the disk (see Problem 9.3). The factor of 1/2reflects the fact that some of the atoms are near the center andsome are far from the center; the factor of 1/2 is an average ofthe square distances. A uniform-density disk whose mass is 15 kgand radius is 0.15 m makes one complete rotation every 0.3 s.

(a) What is the moment of inertia of this disk?
I = kgÂ·m2
(b) What is its rotational kinetic energy?
Krot = J
(c) What is the magnitude of its rotational angular momentum?
Lrot = kgÂ·m2/s

The motion of spinning a hula hoop around one's hips can be modeled as a hoop rotating around an axis not through the center, but offset from the center by an amount , where is less than , the radius of the hoop. Suppose Maria spins a hula hoop with a mass of and a radius of around her waist. The rotation axis is perpendicular to the plane of the hoop, but approximately from the center of the hoop. a. What is the rotational inertia of the hoop in this case? b. If the hula hoop is rotating with an angular speed of , what is its rotational kinetic energy?