A 2.50kg grinding wheel is in the form of a solid cylinder of radius 0.110m . Pa

Question

A 2.50kg grinding wheel is in the form of a solid cylinder of radius 0.110m .

Part A

What constant torque will bring it from rest to an angular speed of 1700rev/min in 2.40s ?

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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(?2??1)=?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 1700rev/min ?

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A 2.50kg grinding wheel is in the form of a solid cylinder of radius 0.110m .

Part A

What constant torque will bring it from rest to an angular speed of 1700rev/min in 2.40s ?

? =   N?m  

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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(?2??1)=?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 1700rev/min ?

K =   J  

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Explanation / Answer

m=2.5 kg

r=0.11m

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

=2.5*0.11^2/2

=0.015125 kg.m^2

=15.125*10^-3 kg.m^2

A)

Torque T=I*alpa

here,

alpa=w/t

=(1700*2pi/60)/2.4

=74.14 rad/sec^2

now,

T=I*alpa

=15.125*10^-3*(74.14)

=1.121 N.m

B)

theta=1/2*alpa*t^2

=1/2*(74.14)*2.4^2

= 213 rad

=33.92 rev

C)

w=T*theta

=1.121*213

=248.77 J

D)

W=1/2*I*w^2

=1/2*(15.125*10^-3)*(74.14)^2

=41.57 J

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