****Please indicate final answers clearly ****** A 1.70 kg grinding wheel is in
ID: 1489423 • Letter: #
Question
****Please indicate final answers clearly ******
A 1.70 kg grinding wheel is in the form of a solid cylinder of radius 0.160 m. What constant torque will bring it from rest to an angular speed of 1000 rev/min in 2.40 s? Through what angle has it turned during that time? Use equation W = tau_z (theta_2 - theta_1) = tau_z Delta theta to calculate the work done by the torque. What is the grinding wheel's kinetic energy when it is rotating at 1000 rev/min? Compare your answer in part (D) to the result in part (C).Explanation / Answer
a) angular speed w = 1000 x 2pi rad / 60s = 104.72 rad/s
using wf = wi + alpha*t
0 = 104.72 - alpha*2.40
alpha = 43.63 rad/s^2
torque = I x alpha = (1.70 x 0.160^2 / 2) x 43.63 = 0.95 Nm
b) using wf^2 - wi^2 =2 x alpha x theta
0 - 104.72^2 = 2* 43.63 * theta
theta = 125.67 rad
c) Work done = 0.95 ( 125.67) = 119.39 J
d) KE = Iw^2 / 2
= (1.70 x 0.16^2 / 2) x 104.72^2 /2 = 119.32 J
e) yes the results are same.
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