Consider a potter’s wheel with a mass of 3.3 kg and a diameter of 0.40 meters. S

Question

Consider a potter’s wheel with a mass of 3.3 kg and a diameter of 0.40 meters. Suppose a lump of clay with a mass of 0.7 kg is placed 0.15 meters from the center of the wheel as shown.

1. If you model the wheel as a disk and the clay as having a very small size, what is the moment of inertia of this system?

2. Two forces F1 = 2.5 N and F2 = 3.5 N are applied as shown in the figure.

a. Will the torque from F1 produce clockwise or counterclockwise rotation?

b. Will the torque from F2 produce clockwise or counterclockwise rotation?

c. Calculate the net torque on the potter’s wheel

3. Calculate the angular acceleration of the wheel with the clay on it.

4. If the wheel starts at rest and the two forces are applied over a time of 2.3 seconds, what will be the angular velocity of the wheel?

Explanation / Answer

mass of the disk, m1=3.3 kg

diameter of the disk, d=0.4m

radius of the disk, r1=0.2 m

mass of the clay, m2=0.7 kg

clay position from the center of the disk is r2=0.15 m

1)

moment of inertia I=1/2*m1*r1^2+m2*r2^2

I=(1/2*3.3*0.2^2)+(0.7*0.15^2)

I=0.08175 kg.m^2

I=81.75*10^-3 kg.m^2

2)

force F1=2.5N at 35 degrees

force F2=3.5N at 75 degrees

a)

F1 produce the counterclockwise torque

b)

F2 produce the clockwise torque

c)

net torque T=(r1XF2)-(r1XF1)

=(r1*F2*sin(75))-(r1*F1*sin(35))

=(0.2*3.5*sin(75))-(0.2*2.5*sin(35))

=0.389 N.m

3)

torque=I*alpa

0.389=81.75*10^-3*alpa

==> alpa=4.76 rad/sec^2

angular acceleration, alpa=4.76 rad/sec^2

4)

time t=2.3 sec

angular velocity w=wo*alpa*t

w=0+4.76*2.3

w=10.95 rad/sec

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