Question

A 2.00 kg grinding wheel is in the form of a solid cylinder of radius 0.130 m . A-What constant torque will bring it from rest to an angular speed of 1000 rev/min in 2.90 s ? B-Through what angle has it turned during that time? C-Through what angle has it turned during that time? Use equation W=τz(θ2−θ1)=τzΔθ to calculate the work done by the torque. D-What is the grinding wheel's kinetic energy when it is rotating at 1000 rev/min ?

Answer 1

A)

moment of inertia is I = 0.5*M*R^2 = 0.5*2*0.13*0.13 = 0.0169 kg-m^2

Totque T = I*alpha

alpha is the angular accelaration

then using wf = wi + (alpha*t)

wi = initial angular speed = 0 rad/s

wf = 1000 rev/min = 1000*2*3.142/60 = 104.73 rad/s

alpha = (wf-wi)/t = (104.73-0)/2.9 = 36.12 rad/s^2

then Torque is T = I*alpha = 0.0169*36.12 = 0.61 N-m

B) theta = (wi*t) +(0.5*alpha*t^2) = (0)+(0.5*36.12*2.9*2.9) = 151.8846 rad

C) Work done by torque is W = T*(theta) = 0.61*151.88= 92.64 J

D) according to Work energy theorem

W = change in KE = Kf-Ki

Ki = 0 J

then Kf = W = 92.64 J