A cylindrical rod 24.0 cm long with mass 1.20 kg and radius 1.50 cm has a ball of diameter 8.00

A cylindrical rod 24.0 cm long with mass 1.20 kg and radius...

A cylindrical rod 24.0 $\mathrm{cm}$ long with mass 1.20 $\mathrm{kg}$ and radius 1.50 $\mathrm{cm}$ has a ball of diameter 8.00 $\mathrm{cm}$ and mass 2.00 $\mathrm{kg}$ attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The system is free to pivot about the bottom end of the rod after being given a slight nudge. (a) After the rod rotates through $90^{\circ},$ what is its rotational kinetic energy? (b) What is the angular speed of the rod and ball? (c) What is the linear speed of the ball? (d) How does this speed compare with the speed if the ball had fallen freely through the same distance of 28 $\mathrm{cm}$ ?

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is the energy stored by an object due to its position within a gravitational field, typically given as mgh. In dynamics problems, especially those involving rotations or pendulum swings, the loss in gravitational potential energy is a key source of the kinetic energy that the object acquires as it moves.

Rotational Kinetic Energy

Rotational kinetic energy is the energy associated with the rotation of an object about an axis. It is given by the expression (1/2)I?², where I represents the moment of inertia and ? is the angular speed. This concept is central in analyzing energy changes in systems where objects rotate as they convert potential energy into kinetic energy.

Moment of Inertia

The moment of inertia quantifies an object’s resistance to angular acceleration about a given axis. It depends on the mass distribution relative to the axis of rotation. Calculating the moment of inertia for complex or composite systems often requires integrating over their mass distributions or summing up the contributions of different components.

Conservation of Mechanical Energy

The principle of conservation of mechanical energy states that in an isolated system, the total energy (potential plus kinetic, including both translational and rotational forms) remains constant. This concept is used to relate the gravitational potential energy lost during motion to the kinetic energies (both rotational and linear) gained by the system.

Angular and Linear Relationships

In rotational motion, there is a direct relationship between the angular speed of an object and the linear speed of a point at a particular radius from the axis of rotation. This relationship is given by the formula v = ?r, which is essential for converting between the rotational motion of the system and any translational motion experienced at the periphery.

Transcript

00:01 In this exercise we have a cylindrical rod that has a mass of 1 .2 kilograms, a length of 24 centimeters, and a radius of 1 .5 centimeters.

00:12 On the top of the rod, there is a ball that has a mass of 2 kilograms and a radius of 4 centimeters.

00:23 And initially, the system is positioned vertically and is at rest.

00:30 In question a, we have to suppose that the system rotates downwards and we have to find what is the kinetic energy of the system when it reaches the horizontal position.

00:50 That is when it is at an angle of 90 degrees from the original position.

00:57 So basically what we have to do is to use the conservation of energy principle.

01:06 So we know that the kinetic energy plus the potential energy initially is equal to the final kinetic energy plus the final potential energy.

01:21 I'm going to choose the zero of the potential energy to be at the end of the rod around which the rod rotates.

01:31 So i'm going to choose to be at this green dot right here.

01:37 So the final, i'm sorry, the initial potential energy, actually the final potential energy is zero.

01:46 And since the system starts at rest, the initial kinetic energy is zero.

01:51 So i know that the final kinetic energy will be equal to the initial potential energy.

01:58 Now the initial potential energy, i know that the initial potential energy is equal.

02:06 To the potential energy of the rod, and that is the mass of the rod, m, times g, times the position of the center of mass with respect to the axis of rotation, so that's l over 2, plus the kinetic energy of the ball, and that is m times the, times the gravitation acceleration, g times l plus r.

02:43 So this is the final kinetic energy.

02:47 If we want to find the numbers, so we have m.

02:55 M has a value of 1 .2 kilograms times g, which is 9 .8 meters per second squared times l, that is 24 centimeters, divided by 2 plus capital m, which is 2, kilograms times 9 .8 meters per second squared times 0 .24 meters plus 0 .04 meters plus 0 .04 meters...

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