A 2.75-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. P

Question

A 2.75-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. Part A What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s? Express your answer with the appropriate units. Value Units My Answers Give Up Part B Through what angle has it turned during that time? My Answers Give Up Part C Use equation W = T_2 (theta_2 - theta_1) = t_1 delta theta to calculate the work done by the torque. Express your answer with the appropriate units. Value Units My Answers Give Up

Explanation / Answer

A.

torque = I*alpha

wf = wi + alpha*t

alpha = (wf - wi)/t

torque = I*(wf - wi)/t

I = 0.5*m*r^2 = 0.5*2.75*0.10^2 = 0.01375 kg-m^2

1200 rpm = 125.66 rad/sec

Torque = 0.01375*(125.66 - 0)/2.5 = 0.691 N-m

B. Angle = w)avg.*dt

average w = (1200 + 0)/2 = 600 rpm = 62.83 rad/sec

angle = 62.83*2.5 = 157.075 rad.

C. W = torque*d(theta)

W = 0.691*157.075 = 108.54 J

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