Question

A grinding wheel is a uniform cylinder with a radius of 8.50 cm  and a mass of 0.480 kg . A) Calculate its moment of inertia about its center. A is I = 1.73×10⁻³ kg⋅m2  . B) Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 5.10 s . Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 53.0 s . I tried doing B, but it keeps saying it's wrong.  

         

Answer 1

Dear student, please refer to the question below.
Question:

A grinding wheel is a uniform cylinder with a radius of 7.20 cm and a mass of 0.470 kg .

A) Calculate its moment of inertia about its center.

-Answer is 1.22 x 10^-3 kg*m^2

B) Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 5.10 s . Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 49.0 s .

Please help with part B, Thanks!

Ans:

A) we can consider grinding wheel as solid disk with uniform density.

then moment of inertia I = m r^2 / 2

I = 0.470 x (0.072^2 ) /2 = 1.22 x 10^-3 kg m^2

B) torque due to frictional force = I x alpha

angular acc due to friction = (wf - wi) / t

= ( 0 - (1500 x 2pi rad/60sec) ) / 49

= 157.08 / 49 =3.21 rad/s^2

torque = - 1.22 x 10^-3 x 3.21 =-3.91 x 10^-3 Nm

now it accelerates from o to 1750 rpm in 5.10 sec

alpha = ((1750x 2pi rad/60sec) - 0 ) / 5.10 = 35.93 rad/s^2

torque due to force + torque due to friction = I x alpha

torque - 3.91 x 10^-3 = 1.22 x 10^-3 x 35.93

torque = 0.0477 Nm