The drawing shows two identical systems of objects; each...

The drawing shows two identical systems of objects; each consists of the same three small balls connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at a different place, as shown. The same force of magnitude $F$ is applied to the same ball in each system (see the drawing). The masses of the balls are $m_{1}=9.00 \mathrm{kg}, m_{2}=6.00 \mathrm{kg}$, and $m_{3}=7.00 \mathrm{kg} .$ The magnitude of the force is $F=424 \mathrm{N}$. (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b) Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force moves with the system and always points along the $4.00-\mathrm{m}$ rod. What is the angular velocity of each system after $5.00 \mathrm{s} ?$

Key Concepts

Rotational Kinematics

Rotational kinematics deals with the relationships between angular displacement, angular velocity, and angular acceleration. When an object starts from rest with a constant angular acceleration, its angular velocity after a certain time can be determined using kinematic equations analogous to those in linear motion, such as ? = ?t. This concept is vital for predicting the state of a rotating system after a given period under a known angular acceleration.

Torque

Torque is the measure of the force's ability to produce rotation about an axis. It is defined as the cross product of the lever arm (the distance from the axis to the point of force application) and the force applied. The magnitude of the torque is given by the product of the force magnitude, the distance, and the sine of the angle between the force direction and the lever arm. Understanding torque is essential when analyzing rotational effects in a system.

Moment of Inertia

The moment of inertia is the rotational analogue of mass in linear motion. It measures how much an object resists angular acceleration about a given axis. For systems of discrete masses, it is calculated by summing each mass multiplied by the square of its distance from the axis. This concept is crucial when comparing systems with different distributions of mass relative to the axis of rotation.

Newton's Second Law for Rotation

Newton's second law for rotational motion 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 allows us to determine how quickly a rotating object will speed up or slow down when subjected to a net torque. It plays a central role in linking the applied force (through torque) to the resulting rotational motion.

Transcript

00:01 Hello, my name is david.

00:02 In this video, we'll cover the relationship between moment of inertia, torque, and angular velocity.

00:07 So for this problem, we have two similar systems, but with different axis of rotation.

00:14 And we're applying the same 4 of 424 newtons pointing along the 4 meter rod, where mass 1 is equal to 9 kilograms, mass 2 is equal to 6 kilograms, and mass 3 is equal to 7 kilograms.

00:33 Now for part a we want to find the moment of inertia for each of the systems.

00:43 So we must recall that the moment of inertia is equal to the sum of each of the masses times the distance from each mass to the axis of rotation square.

00:55 So for system a, the moment of inertia is going to be equal to mass 1 times r1 square plus mass 2 times r2 square plus mass 3 times r3 squared.

01:13 Now for this system we could see that mass 1 is a distance of zero meters from the axis of rotation so the first term is going to become zero.

01:23 So we had that the moment of inertia is actually m2 r2 squared plus m3r3 square so we could plug in our values now so mass 2 is 6 kilograms and the distance from it to the axis of rotation is 3 meters and then the third mass is 7 kilograms and the distance from it to the axis of rotation is 5 meters so for system a we get 54 kilogram meter square plus 175 kilograms meter square which gives us a moment of inertia of 229 kilograms meter square now similarly for system system b the moment of inertia is equal to mass 1 r1 square plus m2 r2 square plus m3 r3 square now for this system we could see that the distance from mass 3 to the axis of rotation is 0 meters so the last term is going to become 0 so the moment of inertia for system b is going to be m1 r1 square plus m2 r2 square so mass 1 again is 9 kilograms and the distance from it to the axis of rotation is 5 meters plus mass 2 which is 6 kilograms and the distance from it to the axis of rotation is 4 meters in this case so we get 225 kilograms meter square plus 96 kilograms meter square which gives us 321 kilograms transmitter square.

04:45 Now for part b we want to find the torque or the net torque in each of the systems.

04:51 So we must recall that the net torque is equal to the force times the distance from the force to the axis of rotation.

05:03 So let's label recording the system for torque and let the direction counterclockwise be positive.

05:18 So for system a, we get that the net torque is equal to to negative of the force times the distance from it to the axis of rotation.

05:32 So the distance is 3 meters.

05:35 So we get negative 424 newtons times 3 meters, which gives us negative 1272 newton's meter.

05:53 But now since the information given was we're up to three significant figures, we want to round to 3 significant figures.

06:01 So the net torque consistent may is equal to negative 1270 newton's meter...

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