Rotational kinematics (mass wrapped around inner hub)

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SUMMARY

The discussion focuses on calculating the angular acceleration of a bicycle wheel with a mass of 6.55 kg and a radius of 38.0 cm, with a mass M of 1.85 kg attached to a string wrapped around an inner hub of radius 5.40 cm. The correct formula for angular acceleration is derived as alpha = Mg*r / (MR^2 + Mr^2), leading to an angular acceleration of 1.06 rad/s². The initial incorrect approach used the tension at the outer radius instead of the inner radius, which clarified the dependency on the inner radius in the calculations.

PREREQUISITES
  • Understanding of rotational dynamics and angular acceleration
  • Familiarity with torque and its calculation (torque = FR)
  • Knowledge of moment of inertia for a ring (I = mr²)
  • Basic principles of Newton's second law (F = Ma)
NEXT STEPS
  • Study the relationship between linear and angular acceleration in rotational systems
  • Learn about the effects of friction in rotational dynamics
  • Explore the concept of work done by external torque in mechanical systems
  • Investigate the applications of rotational kinematics in real-world scenarios, such as bicycles and machinery
USEFUL FOR

Students studying physics, particularly those focusing on rotational dynamics, as well as educators and anyone involved in mechanical engineering or related fields.

SamMarine
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Homework Statement


A bicycle wheel is mounted as in the lab and as shown to the right. This wheel has a mass of
6.55 kg, a radius of R = 38.0 cm and is in the shape of a ring. A mass M = 1.85 kg is attached to
the end of a string which is wrapped around an inner hub which has a radius r = 5.40 cm.

Initially, the mass M is a distance h = 72.0 cm above the floor. [Assume friction is negligible!]
a. What will be the resulting angular acceleration of this wheel?
b. How long will it take for the mass M to reach the floor?
c. What will be the total angular displacement of the wheel during the time in which the mass M
is falling to the floor?
d. How much work was done on the wheel by the external torque as the mass M falls to the floor?

Homework Equations


torque = FR = I * alpha
F = Ma
a = alpha * r

I = mr^2 (told to ignore the spokes of wheel, same as hoop)

The Attempt at a Solution


I tried using T*R = I * alpha (T = tension, I = mR^2 where m is mass of wheel)
and Mg - Ma = T and a = alpha *r

I substituted for T, combined them together to solve for alpha,
got alpha = MgR/(mR^2 + MR^2)
which is 5.795

however answer is 1.06rad/s^2. Can you help?
 
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T*R = I*alpha? If I understand the problem description correctly, the tension T acts at a distance r from the hub, not R.

That would explain why the answer you derived is independent of r while the real answer should depend strongly on r.
 
that did it thanks! Mgr/(MR^2 + Mr^2) got me the answer
 

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