Question

1. When you drop the mass holder, the system rotates at angular acceleration a. Suppose you wrapped the string around a larger diameter pulley and dropped the mass holder. What would happen to a, the angular acceleration of the system, and I, the measured moment of inertia of the system?

Does it increase, decrease, or stay the same?

2. Calculate the moment of Inertia for a metal ring with a mass 4.54 kg and an innter radius of 4.86 cm and an outer radius of 10.83 cm.

3. Suppose m =942 g (including the mass holder) is released from rest. The pulley has diameter d = 4.48 cm of the smallest wheel (note: the string is wrapped around this smallest wheel in the 3-step pulley), and you measure the angular acceleration of the pulley = 0.331 rad/s². Assume

- The system is frictionless.
- The string does not slip on the pulley.
- The string is always perpendicular to the diameter of the pulley.
- g=9.80 m/s²

Calculate the moment of inertia of the system.

Answer 1

1] we have the expression for angular acceleration,a= V²/r [where v is the velocity and r is the radius]

Here we wrapped the string around a larger diameter pulley and dropped the mass holder.Here large diameter ,hence large radius ,thereby angular acceleration decreases,since a~1/r

moment of inertia of the system, I= mr² ,So when radius increases,Moment of inertia also increases.

2] Given:-

m= 4.54 kg

r_{i =}4.86cm

r₀ =10.83 cm.

I = m [r₀²_-r_i²] =4.54 [10.83²-4.86²]*10^-4 = 0.0425kgm²

3] Moment of inertia of the system,I=mr² = 0.942*[2.24*10^-2]²=2.11*10^-4 kgm²