Question

at With the meter stick on the support stand, suspend a gram mass m = 100g the 15-cm mark on the meter stick. Then adjust the lever arm for a gram mass m2 = 200g on the other side of the axis. See Figure. X axis X X2 m m2 Record the mass and position x as read on the meter stick and then record the lever arms. Compute the torques and find a percent difference between the clockwise (Tow) torques and counterclockwise (1.c) torques.

Answer 1

M₁ = 100g

M₂= 200g

Mass of meter stick is 87g

As meter stick is placed on the stand Let us divide the Scale into two halfs

Mass of scale per unit length = 87/100 = 0.87g/cm

Length of scale on Left hand side = 50.5cm as stand is located at 50.5cm

M_Lhs = Mass of meter scale on Left Hand Side = length of scale on left hand side multiplied by mass per unit length =  50.5 x 0.87 = 43.935g

length of scale on Right hand side = 100-50.5 =49.5cm

M_Rhs= Mass of meter scale on right hand side = 49.5 x 0.87 = 43.065 (or 87 - 43.935)

M₁ is located at a distance of 35.5cm from balance point let this be R₁ (50.5-15)

R₁= 35.5 cm

Forces on Left hand side are M₁g and M_Lhsg

Torque Lhs = M₁g(R₁) + M_Lhsg (50.5/2) = 100(35.5)g + 43.935(25.25)g = 4659.36g

T_cc = 4659.36g (take g in cm/s²)

M2 is located at a distance of 18cm from balance point let this be R2 (68.5-50.5)

R₂= 18cm

Forces on the Right hand side are M₂g and M_rhsg

Torque _Rhs = M₂g(R₂) + M_Rhsg(49.5/2) = 200(18)g + 43.065 (24.75)g = 4665.86g

T_cw = 4665.86g

Percentage difference is given by (Difference / Average )*100

$\frac{T_{cc}-T_{cw}}{\frac{T_{cc} +T_{cw}}{2}} * 100$

Percentage difference = 0.139407%difference = 0.14%