Question

Main Menu Contents Grades Timer Notes Evaluate Feedback Print Info Course Contents ... w1 3 10.14T - Dynamics of Rotational Motion Rotational Inertia Suppose you exert a force of 212 N tangential to a 0.290-m-radius 89.0-kg grindstone (a solid disk). What torque is exerted? Su Tes /10 What is the angular acceleration assuming neglige opposing friction? Submit A Tries 0/10 What is the angular acceleration if there is an opposing frictional force of 16.9 N exerted 1.41 cm from the ai? Sun Thies 0/10 This discussion is closed. Send Feedback

Answer 1

1.

Torque = Force *distance

$\tau =F*r$

$\tau =212N*0.290m$

ANSWER: ${\color{Red} \tau =61.48Nm}$

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2.

Moment of inertia of the solid disk, $I=\frac{1}{2}mr^{2}$

$I=0.5*89kg*(0.290m)^{2}$

${\color{Red} I=3.74245kgm^{2}}$

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Net Torque = Moment of inertia * Angular acceleration

$\tau_{net} =I\alpha$

$\frac{\tau }{I}=\alpha$

$\frac{61.48Nm}{3.74245kgm^{2}}=\alpha$

ANSWER: ${\color{Red} \alpha=16.428rad/s^{2}}$

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3.

Net Torque = Moment of inertia * Angular acceleration

$\tau_{net} =I\alpha$

$\tau -\tau _{friction}=I\alpha$

$\frac{\tau -\tau _{friction}}{I}=\alpha$

$\frac{\tau -F_{friction}*r}{I}=\alpha$

$\frac{61.48Nm-16.9N*0.0141m}{3.74245kgm^{2}}=\alpha$

$\frac{61.48-16.9*0.0141}{3.74245}=\alpha$

ANSWER: ${\color{Red} \alpha=16.364rad/s^{2}}$

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