Exercise 10.33 A 2.30 kg grinding wheel is in the form of a solid cylinder of ra

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Exercise 10.33

A 2.30 kg grinding wheel is in the form of a solid cylinder of radius 0.140 m .

Part A

What constant torque will bring it from rest to an angular speed of 1600 rev/min in 2.50 s ?

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Part B

Through what angle has it turned during that time?

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Part C

Use equation W=z(21)=z to calculate the work done by the torque.

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Part D

What is the grinding wheel's kinetic energy when it is rotating at 1600 rev/min ?

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Exercise 10.33

A 2.30 kg grinding wheel is in the form of a solid cylinder of radius 0.140 m .

Part A

What constant torque will bring it from rest to an angular speed of 1600 rev/min in 2.50 s ?

=   Nm  

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Part B

Through what angle has it turned during that time?

=   rad  

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Part C

Use equation W=z(21)=z to calculate the work done by the torque.

W =   J  

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Part D

What is the grinding wheel's kinetic energy when it is rotating at 1600 rev/min ?

K =   J  

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Explanation / Answer

torque = I*alpha

alpha = wf -wi /t

wf = 1600 rev/min = 167.55 rad/s

alpha = 167.55 - 0 /2.5 = 67.02 rd/s^2

I = inetria of solid cylinder = 1/2 * mr^2 = 0.02254 kg.m^2

torque = 1.51 N-m

part b )

dtheta = 1/2 * alpha*t^2

dtheta = 209.4375 rad

part c )

w = torque * dtheta

w = 1.51 * 209.4375 = 316.25 J

part d )

KE = 1/2 * Iw^2

KE = 1/2 * 0.02254 * (167.55)^2

KE = 316.38 J

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