Physics 151 Rotational Equilibrium Name: Objective: In this laboratory you will study an object in rotational equilibrium. Set up a meter stick with three clamps suspended between two pulleys as shown. Ny Pagelof4 I Physics 151 Rotational Equilibrium * The masses Mass of the ruler mg = 92.0 kg Mass of the clamp ma= 22.0 kg Mass of the clamp mg = 21.8 kg Mass of the clamp mg = 22.7 kg my=___ 90__kg m,=_709____ kg my=_700___ kg Moy =My +mp + me +mg = kg * Paosition of the clamps a= 10 m a b X c b= 40 m } & ' b ' c= 90 m # c - - xgr = center of the ruler's mass only = 50cm + Angles @,=__35_ 8,=_85_ @.=_35" ANALYSIS. Locate the center of mass of the combination of ruler and clamps A, B, C, which is Xcu. This is not the center of mass of the meter stick, xg, which should be at about 50 cm or 0.50 m. Mg *Xpg+ My =X+ Mg *Xg+ M *=X Xew="" " - . - . . s= Mg + My + Mg T Mme Page 20f 4 Physics 151 Rotational Equilibrium Make a scale drawing of your experimental setup. The drawing should be to scale spatially, i.e. correct angles and points of application of the forces. The drawing should also represent all of the forces to scale. a. Set up an x-y coordinate system horizontally and vertically (the direction of gravity). In this coordinate system, find EFx and EFy. Note that at point A, we have FAX = -71 sin 01 and FAy = T1 cos 01. Similarly, at point C, we have Fox = 72 sin 02 and Fey = T2 Cos 02. Also, at point B, we have Fox = 0, FBy = my * g. So now we are ready to do our calculations: For EF, we have FAX + Fox = Our percent error is %4= FallFAX . 100 = IFcl+ FAXd For EFy we have: Fy+ = Fay + Fcy= Fy- = m3 *g + mem *g= and | Fy+1 - |Fy_| = and our percent error is %= 15+ 5- . 100 = Ey+ +Fy-1 b. (Optional) c. Now we calculate the sum of the torques about the support point near the end of the meter stick. Since which end to use is not specified, I recommend using support point A. Then we proceed as follows: dA = 0., de = (XB - XA) * cos es , dcm = (XCM - XA) * cos es , dc = (xc - XA) * cos es Then for our torques we have: T+ = Fcy * dc = T_ = FBy * dg + mcm * g * dcm So now our net torque is Er = It+| - It_| = And our percent error is %4= The * 100 = 2 Page 3 of 4 Physics 151 Rotational Equilibrium d. Now we ask you to calculate the torques about the center of mass of the stick plus clamps. In this case, dem = 0. The other distances need to be recalculated accordingly. Also, which forces produce positive torques (tend to rotate counterclockwise) and negative torques (tend to rotate clockwise) will be different. Thus: dA = dB = deM = 0., de= Therefore: T+ = T. + Er= And the percentage error is %A =