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Test 3 sol

# Test 3 sol - the fulcrum If we imagine that all of the...

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Solutions to Test Three 1)The tangential speed of a spinning object can be found by thinking of the  circumference of the wheel as the “distance” traveled in each rotation; the tangential  speed is therefore the number of circumferences per second, or V T  = (2 pi) (20 m) x (1circumference / 50 sec)  V T  = 2.512 m/s 2)You must sum the torques in this problem, and the torque from the measuring stick is  tricky.  Its mass lies all along its length, not in one location at one distance from the  fulcrum like the rock.  The key is to only worry about the stick's center of mass and  the torque it contributes.   The center of mass of the stick will be at its center, positioned one meter to the right of

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Unformatted text preview: the fulcrum. If we imagine that all of the stick's mass is concentrated there, we'll have an identical problem from a physics standpoint, and it is one we can solve: T = F x R T total = 0 T total = T rock + T stick's Center of Mass T total = (M rock G) x (-1 m) + (M stick G) x (1 m) = 0 M rock = M stick = 30 kg 3)If the person's usual weight on Earth is 600 N, then by F = MA, F = M A 600 N = M (10 m/s^2) M = 60 kg Now, using F = MA again with A = 26.7 m/s^2 for Jupiter, F = (60 kg) (26.7 m/s^2) = 1602 Newtons...
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