I have a ruler with a 50g mass hanging at 10cm and the balance positioned at 15.

Question

I have a ruler with a 50g mass hanging at 10cm and the balance positioned at 15.6cm to equal it out. My reference point is 15.6cm and my distance for the ruler side without the mass is 24.8m with a change in distance at 3.8688m. how do I figure out the torque of the ruler?

Calculate the torque of the ruler about the referencepoint.

Note:   Use neither the mass of theruler nor the weight of theruler; the mass of the

ruler is provided as a reference only in order to compute the percent error in Part C. The purpose of this part of the experiment is to calculate the mass of the ruler.

Explanation / Answer

best question i am attempting to answer uptil now. this question is no doubt concerned with balancing of torque. the hung mass is known. position where it is hung is also known. position of pivot(reference balancing point is also known). so the only thing to be calculated is mass of ruler. since the balance point is 24.8m without the mass, the total length of ruler is 24.8*2=49.6m that is this is a almost a half meter scale. a ruler is obviously a uniformly distributed mass(distributed along length). the center of mass(COM) of such a ruler is at its mid point. if a pivot is put in between the scale for balancing, the ruler may be considered to be of two parts(one on each side of pivot). these two parts will have their COM at mid point of the parts. so now it is easy to calculate the torque. right part puts a clockwise torque while the left part puts an anticlockwise torque. now when the pivot is not at the center (as in the question at 15.6 cm), torque due to the two parts of ruler will be different. this unbalanced torque will be balanced by the torque due to the hanging 50g mass at 10cm. length of first part = 15.6cm COM of first part is at 7.8cm length of second part (49.6-15.6) = 34cm COM of second part is at 15.6+(34/2)=32.6cm let the mass of ruler be 'm' so the equation will be torque by hanging mass = difference of torque by two parts of the ruler 50*(15.6-10)= [ (34/49.6)m*17 ] - [ (15.6/49.6)m*7.8 ] this gives m= 30.434 grams approximately . torque of the ruler is nothing but the difference of torque due to the two parts =0.0028 Nm

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.