A thin, light wire is wrapped around the rim of a wheel as...

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius $0.180 \mathrm{~m}$ and moment of inertia for rotation about the axle of $I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}$. A small block with mass $0.340 \mathrm{~kg}$ is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does $-9.00 \mathrm{~J}$ of work as the block descends $3.00 \mathrm{~m}$. What is the magnitude of the angular velocity of the wheel after the block has descended $3.00 \mathrm{~m} ?$

Key Concepts

Non-Conservative Forces and Work Done by Friction

Non-conservative forces, like friction, do work that dissipates mechanical energy from the system. In analyzing dynamic systems, the work done by friction must be considered as it reduces the total energy available for conversion into kinetic energy, affecting the final speeds or angular velocities computed through energy conservation methods.

Energy Conservation in Mechanical Systems

This concept involves the principle that the total mechanical energy in an isolated system, which includes gravitational potential energy, translational kinetic energy, rotational kinetic energy, and work done by non-conservative forces, remains constant. In problems like this, energy lost from potential energy is transformed into kinetic energy of both the moving mass and the rotating wheel, after accounting for energy losses such as friction.

Work-Energy Theorem

The work-energy theorem states that the net work done on a system is equal to its change in kinetic energy. This theorem is applied by summing the work done by all forces, including non-conservative forces like friction, to determine the final energy and thus the final speeds or angular velocities of the system.

Rotational Kinetic Energy and Moment of Inertia

Rotational kinetic energy is the energy due to the rotation of an object, given by one half times the moment of inertia times the square of the angular velocity. The moment of inertia is a measure of an object's resistance to changes in its rotation and plays a role analogous to mass in linear motion, making it essential for analyzing how energy is stored in rotational systems.

Coupling of Translational and Rotational Motion

In systems where a mass is connected to a rotating object, such as a block attached to a string wrapped around a wheel, the motion of the mass (translational) is linked to the motion of the wheel (rotational). This relationship is typically expressed by equating the linear speed of the mass to the product of the wheel's radius and its angular velocity, allowing one to connect energy and kinematic equations from linear and rotational dynamics.

Transcript

00:01 In this problem, we're looking at our old friend the mass suspended from a pulley.

00:06 The only difference here is that this time we have friction in the mix.

00:12 So we're given the radius of the pulley is 0 .18 meters.

00:19 Whoops.

00:21 The moment of inertia is 0 .48 kilogrammeater squared.

00:26 The mass of the weight is 0 .34 kilograms, and the height it's falling is 3 meters, and the work being done by the friction of the axle is 9 joules.

00:42 So this is just our usual energy equation, except with friction added in there.

00:51 So as usual, e is going to be equal to mgh, that is the end.

00:56 Energy entering the system and that's going to be equal to everywhere the energy goes in the system.

01:03 So we have the kinetic energy of the block, one -half mv squared, plus the kinetic energy of the wheel, one -half moment of inertia omega -squared, and then plus the energy going into friction, which i've called ef, friction energy.

01:25 So the first thing to do here is to start getting like terms.

01:31 And i'm going to right away subtract ef from this side since we know everything over here what we're looking for is v so i'm trying to isolate v here mgh minus ef is equal to one half mv squared plus one half i and then omega will become v over r and then that'll be squared okay and now continue isolating i'll pull out the v squared over 2 and that will leave us with m plus i over r squared and now all i need to do is divide it out v squared is going to be equal to m g i'm sorry 2 m g h minus 2 e f from this 2 that we had to multiply by over m plus i over r squared...

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