Question

A grinding wheel is a uniform cylinder with a radius of 6.60 cm and a mass of 0.530 kg . Calculate its moment of inertia about its center. Calculate the applied torque needed to accelerate it from rest to 1600 rpm in 6.30 s if it is known to slow down from 1600 rpm to rest in 51.0 s .

Answer 1

moment of inertia I = (1/2)*m*r^2 = (1/2)*0.53*0.066^2 = 1.15*10^-3 kg m^2

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during acceleration

initial velcoity w1 = 0

final velocity w2 = 1600 rpm = 1600*(2pi/60) = 167.6 rad/s

alpha _1 = (w2 - w1)/t1 = 167.6/6.3 = 26.6 rad/s^2

net torque Tnet = I*alpha1

Tapplied + Tfriction = I*alpha_1

Tapplied = I*alpha_1 - Tfirction

during deceleration

initial velcoity w1 = 1600 rpm = 1600*(2pi/60) = 167.6 rad/s

final velocity w2 = 0

alpha _2 = (w2 - w1)/t1 = -167.6/51 = -3.29 rad/s^2

fricitonal torque T2 = I*alpha2

applied torque ,Tapplied = I*alpha_1 - Tfirction

Tapplied = I*(alpha_1 - alph2)

Tapplied = 1.15*10^-3*(26.6+3.29) = 0.0344 Nm