Can you please answer the following questions: 1) 2) A grinding wheel is In the
ID: 1289997 • Letter: C
Question
Can you please answer the following questions:
1)
2)
A grinding wheel is In the form of a uniform solid disk of radius 6.94 cm and mass 1.94 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.605 N . m that the motor exerts on the wheel. (a) How long does the wheel take to reach its final operating speed of 1 170 rev/mm? s (b) Through how many revolutions does it turn while accelerating? rev The four particles shown below are connected by rigid rods of negligible mass where Y1 = 5.40 m. The origin is at the center of the rectangle. The system rotates in the xy plane about the z axis with an angular speed of 5.50 rad/s. (a) Calculate the moment of inertia of the system about the z axis. (b) Calculate the rotational kinetic energy of the system. In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath It. The diameter of the roller is 2.00 m, and, while being driven Into rotation around a fixed axis, its angular position Is expressed as Where theta is in radians and t is in seconds. (a) Find the maximum angular speed of the roller. rad/s (b) What is the maximum tangential speed of a point on the rim of the roller? m/s (c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation? s (d) Through how many rotations has the roller turned between t = 0 and the time found in part (c)? rotationsExplanation / Answer
1) ? = 2.9t^2 - 0.85t^3
a) ? = d? / dt = d/dt(2.9t^2 - 0.85t^3)
? = d(2.9t^2)/dt - d(0.85*t^3)/dt
? = 2.9*dt^2/dt - 0.85*dt^3/dt
? = 2.9*2*t - 0.85*3*t^2
? = 5.8t - 2.55t^2
if ? is maximum
then d? /dt = 0
d5.8t - 2.55t^2 /dt = 0
5.8 - 5.1t = 0
t = 1.14 s
at t = 1.14s ...? is maximum
so ? max= (5.8*1.14)-(2.55*1.14*1.14) = 3.29 rad/s
b) v = r*? max = 1*3.29 = 3.29 m/s
c) alfa = 5.8 - 5.1t = 0
t = 1.14
d) ? = (2.9*1.14*1.14)-(0.85*1.14*1.14*1.14) = 2.5 radians
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r^2 = (x/2)^2 + (y/2)^2 = 2^2 + 2.7^2 = 11.29 m
2) a) I = (2+3+2+4)*r^2 = 124.19 kg m^2
b) KEr = 0.5*I*w^2 = 0.5*124.19*5.5*5.5 = 1878.37375 J
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3) I = 0.5*M*R^2 = 0.5*1.94*0.0694*0.0694 = 0.0046718692 kg m^2
Torque = I*alfa
alfa = T/I = 0.605/0.0046718692 = 129.49 rad/s^2
wf = (1170*2*3.14)/60 = 122.46 rad/s
a) alfa = (wf - wi)/t
t = wf/alfa = 122.46/129.49 = 0.945 s
b) t(wf^2 - wi)^2 = 2*alfa*theta
theta = wf^2/alfa = 57.09 radians