Exercise 10.33 2.50 kg grinding wheel is in the form of a solid cylinder of radi

Question

Exercise 10.33 2.50 kg grinding wheel is in the form of a solid cylinder of radius 0.190 m. What constant torque will bring it from rest to an angular speed of 1800 rev/minin 2.00 s? Submit My Answers Give U Part B Through what angle has it turned during that time? rad Submit My Answers Give U Part C Use equation W T2002-01) T24ao calculate the work done by the torque Submit My Answers Give U Part D What is the grinding wheels kinetic energy when it is rotating at 1800 rev/min Submit My Answers Give U Part E Compare your answer in part (D) to the result in part (C) The results are the same. The results are not the same.

Explanation / Answer

here ,

m = 2.5 kg

radius , r = 0.19 m

moment of inertia , I = 0.5 mr^2

I = 0.5 * 2.5 * 0.19^2

I = 0.0451 Kg.m^2

a)

angular speed , Wf = 1800 rev / min

Wf = 1800 * 2* pi /60

Wf = 188.49 rad/s

initial angular speed ,Wi = 0

time , t = 2 s

let angular accelration be a

using first equation of motion

Wf = Wi + a * t

a = 94.24 rad/s^2

let the constant torque is T ,

using second law of motion

I*a = T

T = 0.0451 * 94.24

T = 4.25 N.m

the constant torque is 4.25 N.m

b)

angle = 0.5 * a*t^2

angle = 0.5 * 94.24 * 2^2

angle = 188 .48 rad

c)

work done = torque * angle

work done = 4.25 * 188.49

work done is 801.16 J

d)

kinetic energy = 0.5 * I * Wf^2

kinetic energy = 0.5 * 0.0451 * 188.49^2

kinetic energy is 801.16 J

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