A 1.10-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. (
ID: 1494964 • Letter: A
Question
A 1.10-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. (a) What constant torque will bring it from rest to an angular speed of 1350 rev/min in 1.5 s?N · m
(b) Through what angle has it turned during that time?
rad
(c) Use Eq. (10.21) to calculate the work done by the torque. W = z(2 - 1) = z (work done by a constant torque) (10.21) J
(d) What is the grinding wheel's kinetic energy when it is rotating at 1350 rev/min?
J
Compare your answer to the result in part (c). The answer to part (d) is greater.The answer to part (c) is greater. The two answers are the same. A 1.10-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. (a) What constant torque will bring it from rest to an angular speed of 1350 rev/min in 1.5 s?
N · m
(b) Through what angle has it turned during that time?
rad
(c) Use Eq. (10.21) to calculate the work done by the torque. W = z(2 - 1) = z (work done by a constant torque) (10.21) J
(d) What is the grinding wheel's kinetic energy when it is rotating at 1350 rev/min?
J
Compare your answer to the result in part (c). The answer to part (d) is greater.The answer to part (c) is greater. The two answers are the same. (a) What constant torque will bring it from rest to an angular speed of 1350 rev/min in 1.5 s?
N · m
(b) Through what angle has it turned during that time?
rad
(c) Use Eq. (10.21) to calculate the work done by the torque. W = z(2 - 1) = z (work done by a constant torque) (10.21) J
(d) What is the grinding wheel's kinetic energy when it is rotating at 1350 rev/min?
J
Compare your answer to the result in part (c). The answer to part (d) is greater.The answer to part (c) is greater. The two answers are the same. W = z(2 - 1) = z (work done by a constant torque) (10.21) The answer to part (d) is greater.The answer to part (c) is greater. The two answers are the same.
Explanation / Answer
A) Torque (f) = Inertia (I) x angular acceleration (c)
angular speed (w) = 1350rpm x 2pi radians/ 60 seconds = 45pi rad/second
w (final) = w (initial) + (c) x time (t)
45pi= 0 - c(1.5seconds)
c= 30pi = -94.24 rad/s^2
Moment of Inertia of solid cylinder = I = mr^2 / 2
m = 10 kg, r = 0.5 m
I = mr^2 / 2 = 10*0.5^2 / 2 = 1.25
f= I x c
f= 1.25 x 94.24
f= 117.84 N/m
constant torque required is 117.84 N/m
B) Acceleration is -ve from t=0 to t=1.5s
so angle turned from t=0 to 1.5s is s = u*t - 0.5*a*t^2
s = (141.37*1.5) - 0.5*94.24*(1.5)^2 = 106 rad
Total angle from t=0 to rest = 106 rad
C) Work done by the torque
W = (2 - 1) = = 117.84 N/m * 106 rad = 12491.2 J
D) kinetic energy (Ek) = 1/2 * I x w^2 = 0.5*1.25*(45pi)^2 = 12491.2 J
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