The mass of an extended object (something long, like a meter stick) can be found by attaching a mass near one end of the object and then balancing the extended object at its pivot point. When the extended object is in equilibrium, the sum all clockwise and counter clockwise torques must be zero. Therefore, the sum of clockwise torques must be equal to the sum of counter clockwise torques.
If a 0.25kg mass is attached at the 0.15m mark of a meter stick, and the meter stick is then balanced when the pivot point is at the 0.42m point, calculate the mass of the meter stick. Assume you have fou

Question

The mass of an extended object (something long, like a meter stick) can be found by attaching a mass near one end of the object and then balancing the extended object at its pivot point.  When the extended object is in equilibrium, the sum all clockwise and counter clockwise torques must be zero.  Therefore, the sum of clockwise torques must be equal to the sum of counter clockwise torques.
If a 0.25kg mass is attached at the 0.15m mark of a meter stick, and the meter stick is then balanced when the pivot point is at the 0.42m point, calculate the mass of the meter stick.  Assume you have fou

anonymous · Answer

Assume you have found the center of mass to be at the 0.51m mark.  Remember that L1 and L2 are the distances between the points above, not the points themselves.  Note that “g” is in all terms and can therefore be divided out.  Show all your work here (including equation and units). Mass of meter stick (rod) is:

anonymous · Answer

any help?

anonymous · Answer

mass can be found by osscilating the rod about its end and calculating the time period and using the expression for time period to get mass..

mos1635 · Answer

|dw:1334651303354:dw|