Question

A meterstick (L = 1 m) has a mass of m = 0.19 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

What is the tension in the left string right after the right string is cut?

Answer 1

Given that :

length of the meterstick, L = 1 m

mass of meterstick, m = 0.19 kg

using an equation, we have

F_ext = m a_CM

mg - T₁ = m a_CM

T₁ = m (g - a_CM)                                                                     { eq.1 }

we know that, = I

(mg) d = I (a_CM / d)

a_CM = mg d² / I                                                                    { eq.2 }

where, I = moment of inertia of meterstick = I_CM + m L²

I = (1/12) m L² + m L² (1/12) (0.19 kg) (1 m)² + (0.19 kg) (1 m)²

I = (0.0158 kg.m²) + (0.19 kg.m²)

I = 0.2058 kg.m²

inserting the values in eq.2,

a_CM = (0.19 kg) (9.8 m/s²) (0.25 m)² / (0.2058 kg.m²)

a_CM = 0.565 m/s²

Now, using eq.1 -

T₁ = (0.19 kg) [(9.8 m/s²) - (0.565 m/s²)]

T₁ = (0.19 kg) (9.23 m/s²)

T₁ = 1.75 N