Homework #7 Solutions
1)
Think of the wheel as a circle  then cut the circle open, and lay the entire length
down on the ground.
If the wheel rotates once per second, then the wheel covers
this much ground with each rotation.
So,
Linear speed = circumference x angular speed
(2m) x (1 revolution per second) = 2 m/s
2)
The center of mass will also be the balancing point.
Torque = (force) x (arm length)
In order for the bar to remain still, the torques from the 1kg and 3kg weights about the
center of mass (the balance point) must be equal (so they can cancel each other out):
(30N) x (L) = (10N) x (L’)
where L is the distance from the 3kg weight to the center of mass, and L’ is the
distance from the center of mass to the 1kg weight.
From this equation, you can
determine that L’ must be 3 times greater than L, so the balance point must be ¾ of
the length from the 1kg mass.
3)
An easy way to think of this problem is to imagine the second problem turned
upside down.
The car is at the balance point, and now we need to determine the
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 Spring '08
 Turner
 Physics, Force, Work, 2m, 2 m/s, 22 kg, 0.10M, 0.25M

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