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(I) Calculate the net torque about the axle of the wheel shown in Fig. $47 .$ Assume that a friction torque of 0.40 $\mathrm{m} \cdot \mathrm{N}$ opposes the motion.

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$1.4 \mathrm{m} \cdot \mathrm{N},$ clockwise

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

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.

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what is the net torque abo…

we know that here each forces oriented perpendicular to lover arm and we're going to then say that we're gonna call counterclockwise torques to be positive. And we can then say the torque of the applied forces would be equaling 28 Newtons multiplied by 280.4 meters, brother point 24 meters my qualities and then minus 18 Newton's multiplied by 180.24 meters minus 35. Newton's multiplied by 0.12 meters. And so the torque due to the applied forces would be equaling negative 1.8 Newton meters. And so because this torque is, we can say clockwise because it's negative. We're going to assume that the wheel is rotating clockwise and so the frictional torque is counter clockwise. So we can say that the frictional torque is counter clockwise, always opposite of the direction of motion. And so the net torque we can say that to work net would be equaling the torque applied minus. We're in this case, plus the frictional tour, and this would be equaling negative 1.8 Newton meters plus 0.4 new 10 meters, and this is equaling negative 1.4 Newton meters, so we can say that this would be equal to 1.4 Newton meters and this would be clockwise, so the magnitude of the network would be equal to 1.4 Newton meters clockwise. This would be our final answer. That is the end of the solution. Thank you for

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