Question

An green hoop with mass m_h = 2.8 kg and radius R_h = 0.12 m hangs from a string that goes over a blue solid disk pulley with mass m_d = 2 kg and radius R_d = 0.09 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass m_s = 4.1 kg and radius R_s = 0.17 m. The system is released from rest.

5)

What is the tension in the string between the sphere and disk pulley?

8)

What is the magnitude of the velocity of the green hoop after it has dropped 1.53 m?

9)

What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.53 m)?

Answer 1

the only force on the system is the weight of the hoop
F net = 2.8kg*9.81m/s^2 = 27.468 N

The inert masses that must be accelerated are the ring, pulley, and the rolling sphere.

The mass equivalent of M the pulley is found by
torque τ = F*R = I*α = I*a/R
F = M*a = I*a/R^2 -->
M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m

The mass equivalent of the rolling sphere is found by:
the sphere rotates around the contact point with the table.
So using the theorem of parallel axes, the moment of inertia of the sphere
is I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2
M = 7/5*m

the acceleration is then
a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4.1)

= 2.879m/s^2

5) tension between pulley and sphere

= M*a
  
= 7/5*4.1*2.879

= 16.525 N

8) v = √(2as) = √(2*2.879*1.53) = 2.968 m/s

9) ω = v/R =2.968 /0.17 =17.45 rad/s