Consider the system shown in Fig. E9.49. The suspended block...

Consider the system shown in Fig. E9.49. The suspended block has mass $1.50 \mathrm{~kg}$. The system is released from rest and the block descends as the wheel rotates on a frictionless axle. As the wheel is ro- tating, the tension in the light wire is $9.00 \mathrm{~N}$. What is the kinetic energy of the wheel $2.00 \mathrm{~s}$ after the system is released?

Consider the system shown in Fig. E9.49. The suspended block has mass $1.50 \mathrm{~kg}$. The system is released from rest and the block descends as the wheel rotates on a frictionless axle. As the wheel is ro-
tating, the tension in the light wire is $9.00 \mathrm{~N}$. What is the kinetic energy of the wheel $2.00 \mathrm{~s}$ after the system is released?

Key Concepts

Rotational Kinetic Energy

This concept deals with the energy possessed by an object due to its rotation. It is calculated using the formula K = ½ I ?², where I is the moment of inertia and ? is the angular velocity. Understanding this concept is crucial for analyzing how energy is distributed in rotating systems.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to angular acceleration about a rotational axis. It depends on both the mass of the object and the distribution of that mass relative to the axis of rotation. It plays a role analogous to mass in linear motion.

Torque

Torque is the measure of a force’s ability to produce angular acceleration about an axis. It is calculated as the product of the force and the lever arm (the perpendicular distance from the axis of rotation to the line of action of the force). Understanding torque is essential in analyzing rotational dynamics.

Newton’s Second Law for Rotation

This law states that the net torque acting on a system is equal to the product of its moment of inertia and its angular acceleration (? = I?). This principle is foundational for linking forces (or torques) to the resulting rotational motion in a system.

Rotational Kinematics

Rotational kinematics involves the equations that relate angular displacement, angular velocity, angular acceleration, and time. These relations are analogous to linear kinematics and are used to predict the rotational motion of objects undergoing constant angular acceleration.

Transcript

00:01 This question here is asking us about this here diagram.

00:05 A mass is attached by a rope to a pulley.

00:08 It is being pulled down by gravity.

00:11 And there is a tension in the rope we are told of nine newtons.

00:15 And the mass of the block is 1 .5 kilograms.

00:19 Then what we have to find is after two seconds, what is the energy in the wheel? so the first thing that we should do here is finish this force diagram that i've started to draw.

00:33 Draw.

00:34 Tension is of course pulling upward on the block from the rope and gravity is going to be pulling down on the block.

00:40 So if we can figure out gravity, we can figure out what the net force on the block is.

00:45 That's fg.

00:46 It's equal to ma or mg in this instance.

00:51 And that will be equal to the mass 1 .5 times g we know is 9 .81.

01:02 And that will be together equal to 14 .715 newtons.

01:10 So then to get the net force fn is equal to these things just minus each other because they're working in different directions, fg minus t which is 14 .715 minus 9 is equal to 5 .715 okay, so now that that's settled, i'll plug that in here.

01:46 14 .715 newtons.

01:50 We can get to work on the problem itself.

01:53 Now, this will be, since we're looking for energy, it only makes sense to turn to our energy equation.

02:02 The energy that's going to be imparted into it is going to be equal to the energy from this block moving down.

02:11 Which is m, g, h, and that is going to be equal to the energy being put into the system, which will be the kinetic energy of the block, plus the kinetic energy of the pulley.

02:24 The block is, of course, one -half mv squared, plus the energy the pulley, one -half, the moment of inertia, times the angular velocity squared.

02:37 But we don't really need to write all this, because what we're looking for is just the kinetic energy, so we don't need to break it down more than that.

02:46 So then the easiest way to do this will probably be to isolate the kinetic energy of the wheel.

02:53 So just pull that out on the left.

02:56 It will be equal to m .g .h.

03:01 Minus one half mv squared...

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