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calculate the torque due to friction acting on the rotor.

calculate the torque due to friction acting on the rotor.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Physics 45 Moment of Inertia At-Home Lab (e) Recall that torque is given by t =rx F, where r is a vector going from the axis of rotation to the point where the force is applied, and F is the force vector. The shaft that the string is wrapped around has a radius of 1.50 cm. Calculate the torque on the rotor due to the tension in the string (in Nm) during the accelerated motion. I=(6,4889)(0. 0/5) O, 00733 TT =1,33V)0-3 Nm (f) We are now going calculate the torque due to friction acting on the rotor. For this part, only consider the data for the constant velocity motion. When the weights fall at constant velocity and the rotor spins at constant velocity, we know that the angular acceleration and linear acceleration are both zero. There are two opposing forces acting on the weights: gravity and tension. There are two opposing toques acting on the rotor: the torque due to the tension in the string and the torque due to friction. Apply EF = Ma to the hanging weights and apply EText = Ia to the rotor. Then solve for tf. %3D IT= 7.33x103 Nm a: 0,022 mis? a= 9.217 rad/s ィこ 0.4884N セ8.1825 し Nm
Transcribed Image Text:Physics 45 Moment of Inertia At-Home Lab (e) Recall that torque is given by t =rx F, where r is a vector going from the axis of rotation to the point where the force is applied, and F is the force vector. The shaft that the string is wrapped around has a radius of 1.50 cm. Calculate the torque on the rotor due to the tension in the string (in Nm) during the accelerated motion. I=(6,4889)(0. 0/5) O, 00733 TT =1,33V)0-3 Nm (f) We are now going calculate the torque due to friction acting on the rotor. For this part, only consider the data for the constant velocity motion. When the weights fall at constant velocity and the rotor spins at constant velocity, we know that the angular acceleration and linear acceleration are both zero. There are two opposing forces acting on the weights: gravity and tension. There are two opposing toques acting on the rotor: the torque due to the tension in the string and the torque due to friction. Apply EF = Ma to the hanging weights and apply EText = Ia to the rotor. Then solve for tf. %3D IT= 7.33x103 Nm a: 0,022 mis? a= 9.217 rad/s ィこ 0.4884N セ8.1825 し Nm
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