Rotational kinematics (mass wrapped around inner hub)

  • Thread starter SamMarine
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
jbriggs444
Science Advisor
Homework Helper
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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.
 
  • #3
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that did it thanks! Mgr/(MR^2 + Mr^2) got me the answer
 

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