Phsyics - Rotational Motion with a Pulley

Question

Two masses (m1 = 5.0 kg and m2 = 2.8 kg) on opposite sides of pulley are released from rest, with m1 at a height of 0.76 m above the floor and m2 on the floor. The system then rotates without slipping and m1 hits the ground with a speed of 1.7 m/s. The pulley has a radius of 12 cm. What is the rotational inertia of the pulley?

My prof said we should be able to solve this problem using both force equations and energy equations.

Answer 1

You can get the acc from the kinematics 2ad = v_f² - v_i² or a = v_f²/(2d) (all givens)

Adding the force balances for the two weights and the torque balance for the pulley in order to remove tensions from the equation:

m₁g - T₁ = m_ia

T₂-m₂g = m₂

(T₁ - T₂)R_P = I_Pα = I_P a/R_P no slip condition and leading to T₁-T₂ = I_Pa/R_P² add them up:

m₁g - m₂g = (m₁ + m₂ + I_P/R_P²)a (Knowing a, you can solve for I_P)

The other way is to realize that the loss of potential energy is = to the net gain in kinetic energy

The E_K of the pulley is 1/2 Iω² or 1/2 I (v²/R_P²)

The overall energy balance is

m₁gd-m₂gd = v²/2 (m₁+m₂+ I_P/R_P²) and solve for I_P

Hope that helps. Take care.