A2.45-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. Pa

Question

A2.45-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. Part A What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 Express your answer with the appropriate units. Part B Through what angle has it turned during that time? Part C Use equation W = tau_z(theta2 - theta1) = tau_zDeltatheta to calculate the work done by the torque. Express your answer with the appropriate units. Part D What is the grinding wheel's kinetic energy when it is rotating at 1200 rev/min? Express your answer with the appropriate units.

Explanation / Answer

A) initial speed, wi = 0

final speed, wf = 1200 rev/min= 1200 x 2pi rad / 60 sec = 125.66 rad/s

t = 2.5 s

using wf = wi + alpha*t

125.66 = 0 + (alpha * 2.5)

alpha = 50.265 rad/s^2

and torque = I x alpha

       = ( m r^2 /2 ) alpha = (2.45 x 0.1^2 /2 ) x 50.265

    = 0.62 N m

B) angle turned, = wi*t + alpha*t^2 /2

      = 0 + (50.265 x 2.5^2 / 2) =157.08 rad

C) W = torque x delta(theta)

      = 0.62 x (157.08 - 0 ) = 97.4 J

d) KE = I w^2 /2

    = (2.45 x 0.1^2 / 2 ) ( 125.66^2) /2 = 96.72 J

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