A 1.70 grinding wheel is in the form of a solid cylinder of radius 0.140 . Part

Question

A 1.70 grinding wheel is in the form of a solid cylinder of radius 0.140 .


Part A
What constant torque will bring it from rest to an angular speed of 1100 in 2.90 ?
=0.662
Correct


Part B
Through what angle has it turned during that time?
=


Try Again; 2 attempts remaining


Part C
Use equation to calculate the work done by the torque.
=


Try Again; 5 attempts remaining


Part D
What is the grinding wheel's kinetic energy when it is rotating at 1100 ?
=111
Correct


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.
Correct

Explanation / Answer

      The mass of the grinding wheel , m = 1.70 kg       The radius of the wheel, r = 0.140 m ------------------------------------------------------------------------       The moment of inertia of the grinding wheel is                 I = (1/2)mr2                    = (1/2)(1.70 kg)(0.140 m)2                    = 0.01666 kg.m2       The angular speed of the wheel is                      = (1100 rev/min)                          = (1100)(2/60) rad/s                          = 115.133 rad/s ----------------------------------------------------------------------- a)       The angular acceleration of the wheel is                   = (115.133 rad/s)/2.90 s                      = 39.7011 rad/s2       The torque is                      = I                         = (0.01666 kg.m2)(39.7011 rad/s2)                         = 0.662 Nm ----------------------------------------------------------------------- b)       Using the rotational motion relations, we have                             = 0t + (1/2)t2                                = 0+ (1/2)(39.7011 rad/s2)(2.90 s)2                                = 166.94 rad                                = 166.94 rad ----------------------------------------------------------------------- c)       The work done is                            W =                                 = (0.662 Nm)(166.94 rad)                                   = 110.516                                 = 111 J ----------------------------------------------------------------------- d)       The rotational kinetic energy is                             K = (1/2)I2                                      = (1/2)(0.01666 kg.m2)(115.133 rad/s)2                                   = 110.419                                 = 111 J ----------------------------------------------------------------------- e)       The results of the part c and d are same.                   

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