Question

To start her lawn mower, Julie pulls on a cord that is wrapped around a pulley. The pulley has a moment of inertia about its central axis of I = 0.520kg?m2 and a radius of 6.00cm . There is an equivalent frictional torque impeding her pull of ?f = 0.410m?N .

A. To accelerate the pulley at ? = 4.70rad/s2 , how much torque does Julie need to apply to the pulley?

B. How much tension must the rope exert?

Answer 1

Answer 2

$\small \small \\ A.) \text{The total torque should be enough to overcome the frictional torque} \vspace{3mm} \\ \text{and to create the angular acceleration} \vspace{3mm} \\ \therefore \ Torque , \tau = \tau _f + I*\alpha = 0.410 + (0.52*4.7) = 2.854 \ N.m \vspace{3mm} \\ B.) \ Torque = Tension*radius \vspace{3mm} \\ \tau = T*r \vspace{3mm} \\ T = \frac{\tau}{r} = \frac{2.854}{0.06} = 47.57 \ N$

Answer 3