Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

(II) Suppose the force $F_{T}$ in the cord hanging from the pulley of Example 9 of "Rotational Motion," Fig. $21,$ is given by the relation $F_{T}=3.00 t-0.20 t^{2}$ (newtons) where $t$ is in seconds. If the pulley starts from rest, what is the linear speed of a point on its rim 8.0 s later? Ignore friction.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

$17 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

03:37

Suppose the force $F_{\ma…

04:44

A block (mass $=2.0 \mathr…

02:04

A block (mass = 2.0 kg) is…

03:09

02:14

A pulley such as that show…

02:02

The force $\vec{F}$ in $4…

03:35

An object of mass $M=$ 12.…

08:05

A $12.0-\mathrm{kg}$ objec…

02:03

A cable passes over a pull…

01:44

A pulley, with a rotationa…

06:29

Newton's Second Law f…

07:12

Motor $M$ exerts a constan…

01:19

A pulley of radius $2 \mat…

02:16

The force $\vec{F}$ in Fig…

00:41

A cord with negligible mas…

We know that the applied force causes a torque which gives the pulley and angular toleration. We know that the applied force Berries with time. And if this is a case, so will the angular acceleration. We can say that, Ah, in order to find the angular velocity, we can integrate the variable acceleration to stall for this. And then we can then say the speed of a point on the rim is the tangential velocity of the rim of the wheel. So we can set this up and say that Ah, Sigma Tau equaling. Ah, the radius times the force tension. And this would be equaling the moment of inertia multiplied by the angular acceleration. So we find that the angular acceleration would simply be equaling the radius times of force. Tension, which is again burying with time. And this would be divided by the moment of inertia from the definition of the angular acceleration in your acceleration is gonna be equaling the derivative of the angular velocity with respect to Time T. So we can then say that we can integrate from Brother Omega initial to Omega final times d omega, and then this would equal the integral from zero to t of the radius times the force tension divided by the moment of inertia d t. And so we know that the moment of the angular velocity rather is equaling the linear velocity divided by our by the Radius. And this would be equaling the moment of the initial angular velocity plus r divided by I multiplied by the integral from zero to t of the force tension DT. Given that the force tension is the only variable that actually depends on time. So we can then say that the initial angular velocity zero so we can eliminate that. And we can then say that the angular velocity would be equaling the radius, divided by the moment of inertia multiplied by the integral from zero to t of the force tension d t. And we're simply going to then, um, we can simply say that the tangential velocity would be equaling. So the final velocity times the radius and so this would be equaling the radius squared, divided by the moment of inertia, Time of the integral from zero to t of f sub t. Now this would be equaling 3.0 T minus 0.20 t squared times D t. And so we can then solve and say that this is gonna be the 10 gentle Ah, velocity would be equaling r squared, divided by the moment of inertia multiplied by three over to T square minus 0.20 divided by three t cubed and the units Newton seconds. And we can then say that the velocity, the tangential velocity AT T equaling 8.0 seconds. This would be equaling the radius of 0.330 meters, divided by the moment of inertia to be 0.3 85 kilogram meter squared. And then this would, uh, the radius would be squared on, then multiplied by three over two times 8.0 seconds. Quantity squared minus 0.20 divided by three times 8.0 seconds. Quantity cubed. And then again here the the units here would be new 10 seconds and we find that the tangential velocity AT T equaling 8.0 seconds would be approximately equal to 17.5 meters per second. This would be our final answer. That is the end of the solution. Thank you for one

View More Answers From This Book

Find Another Textbook

Numerade Educator

Simon Fraser University

04:30

MPLE 2.4Circuit Analysis As an example of how the tools p…