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CBSE Class 11-science Answered

A flywheel of mass 25 kg & radius 0.2 m is rotating at 240 rpm. What is the torque necessary to bring it to rest in 20 sec? If the torque is due to a force applied tangentially on the rim of the wheel, what is the magnitude of the force? Assume that mass of flywheel is concentrated at its rim.
Asked by bhawnanaman | 25 Apr, 2016, 06:00: AM
answered-by-expert Expert Answer
begin mathsize 14px style Given space that space the space mass space of space the space flywheel comma space straight M space equals space 25 space kg semicolon Radius space straight R space equals space 0.2 space straight m space semicolon space straight omega subscript straight o space equals space 240 space straight r. straight p. straight m comma space equals space fraction numerator 240 space over denominator 60 cross times 60 end fraction straight r. straight p. straight s. space equals space 4 space straight r. straight p. straight s space equals space 8 straight pi space rad space straight s to the power of negative 1 end exponent space semicolon space straight t space equals space 20 space straight s space and space straight omega space equals space 0 from space relation space straight omega space equals space straight omega subscript straight o space plus space αt 0 space equals space 8 straight pi space plus space straight alpha cross times 20 rightwards arrow straight alpha space equals space minus space 0.4 space πrad space straight s to the power of negative 2 end exponent Moment space of space inertia space of space the space flywheel space is space straight I space equals space MR squared rightwards arrow space straight I space equals space 25 cross times left parenthesis 0.2 right parenthesis squared space equals space 1 space kg space straight m squared Torque space needed space to space bring space the space flywheel space to space rest comma space straight tau space equals space Iα rightwards arrow space straight tau space equals space 1 cross times 0.4 space straight pi space equals space 1.257 space Nm Let space the space force space applied space tangentiallyon space the space rim space of space the space flywheel space be space straight F comma space then straight tau space equals FR or space straight F space equals space straight tau over straight R equals fraction numerator 1.257 over denominator 0.2 end fraction equals 6.285 space straight N space end style
Answered by Yashvanti Jain | 25 Apr, 2016, 11:25: AM

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