You are testing a small flywheel (radius 0.166 m ) that will be used to store a small amount of

You are testing a small flywheel (radius 0.166 m ) that...

You are testing a small flywheel (radius $0.166 \mathrm{~m}$ ) that will be used to store a small amount of energy. The flywheel is pivoted with low-friction bearings about a horizontal shaft through the flywheel's center. A thin, light cord is wrapped multiple times around the rim of the flywheel. Your lab has a device that can apply a specified horizontal force $\overrightarrow{\boldsymbol{F}}$ to the free end of the cord. The device records both the magnitude of that force as a function of the horizontal distance the end of the cord has traveled and the time elapsed since the force was first applied. The flywheel is initially at rest. (a) You start with a test run to determine the flywheel's moment of inertia $I$. The magnitude $F$ of the force is a constant $25.0 \mathrm{~N},$ and the end of the rope moves $8.35 \mathrm{~m}$ in $2.00 \mathrm{~s}$. What is $I$ ? (b) In a second test, the flywheel again starts from rest but the free end of the rope travels $6.00 \mathrm{~m} ;$ Fig. $\mathrm{P} 10.90$ shows the force magnitude $F$ as a function of the distance $d$ that the end of the rope has moved. What is the kinetic energy of the flywheel when $d=6.00 \mathrm{~m} ?$ (c) What is the angular speed of the flywheel, in rev/min, when $d=6.00 \mathrm{~m}$ ?

Key Concepts

Moment of Inertia

This is a measure of an object’s resistance to changes in its rotational motion. It quantifies how the mass of a body is distributed relative to its axis of rotation, playing the same role for rotation as mass does for linear motion. In problems involving rotational dynamics, knowing the moment of inertia is crucial for predicting the angular acceleration produced by a given torque.

Torque

Torque is the rotational equivalent of force. It is calculated as the product of the applied force and the lever arm (the perpendicular distance from the axis of rotation to the line of action of the force). Torque is responsible for causing angular acceleration in a rotating system, serving as the driving factor behind changes in rotational motion.

Work-Energy Principle in Rotational Motion

This principle states that the work done on a rotating object by applied torques is equal to the change in its rotational kinetic energy. When a force is applied over a displacement (or equivalently when a torque is applied over an angular displacement), energy is transferred into the rotational motion of the object, making it key to solving problems involving energy storage and conversion in rotational systems.

Angular Kinematics

Angular kinematics deals with the description of rotational motion using parameters such as angular displacement, angular velocity, and angular acceleration. These concepts are analogous to linear kinematics but account for the rotational nature of the motion, and they are linked by rotational analogues of the equations of motion, which are essential for analyzing the time evolution of systems like flywheels.

Linear and Angular Relationships

This concept bridges the gap between linear and rotational motion. It establishes that the linear displacement along the rim of a rotating object is proportional to its angular displacement, with the proportionality constant being the radius. This relationship allows one to convert between measurements of linear distance and the corresponding angular quantities, which is particularly useful when dealing with forces applied via a cord wound around a rotating object.

Transcript

00:01 In this problem on the topic of dynamics of rotational motion, we have to, we are to test a small flywheel that it's going to be used to store a small amount of energy.

00:10 Now, we want to calculate the moment of inertia of the flywheel.

00:16 We then want to calculate the kinetic energy of the flywheel after a second test, and we are given a graph of force as a function of distance.

00:28 And finally we want to find the angular speed of the flywheel in revs per minute given the distance d.

00:36 Now we know the work done by the force is equal to the kinetic energy gained by the flywheel, and the work is the area under the force versus d graph.

00:48 So for part a, we know that the pull is constant, so the linear and angular accelerations are constant.

00:54 And so hence, fee is equal to two times the average velocity 2av, which is simply two times the distance that the rope moves divided by the time t.

01:10 We also know that the angular velocity omega is equal to v over r, where r is the radius of the flywheel.

01:19 So this we can write as 2 times d over t times r using the expression of v for v.

01:33 Now the work done is equal to the kinetic energy of the flywheel.

01:37 So the work done, f times d, is equal to the angular kinetic energy, a half -i omega -squared, where i is the moment of inertia.

01:51 So we can write this as a half times i and replace omega 2d over tr, and that's all squared.

02:03 So now we can solve for the moment of inertia i as required.

02:07 And we get the moment of inertia i is equal to f t squared times r squared over 2d and so if we substitute our values in here we can find this moment of inertia so the force is constant at 25 newtons the time t is two seconds and that's squared the radius of the flywheel is 0 .166 meters squared divided by 2 times d, which is 2 times 8 .35 meters...

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