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I've been trying to figure out a question that was on our finals but still could not manage to do it. It goes as:

A ball of mass $m$ and radius $r$ (moment of inertia $I_{cm} =(2/5)mr^2$) is placed on the inside of a frictionless circular track of radius R0 as shown in the figure. It starts from rest at the vertical edge of the track, and since there is no friction, it slides down without rotation.

The horizontal section of the track starting from B is a surface with coefficient of kinetic friction µk. If the ball starts to roll without slipping after traveling a distance d, what is the expression for the coefficient of kinetic friction in terms of the given parameters?

system

When I tried solving the question, I'd come to a point where I had Initial Potential Energy = Work done by static friction + Work done by kinetic friction + Rotational kinetic energy.

What I could not manage to do was to express static friction coefficient and angular velocity in terms of the given parameters, ultimately not allowing me to find the answer. How am I supposed to express them?

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  • $\begingroup$ You also need to use $I\dot \omega=r F$ where $F=\mu mg $ is the friction force. Only when $r\omega=v$ will the ball start to roll without slipping. $\endgroup$
    – mike stone
    Feb 22, 2022 at 22:47
  • $\begingroup$ Good point but it is still not enough since I will still have either w or v in the formula, which I need to convert to the given parameters. $\endgroup$
    – KayB
    Feb 22, 2022 at 22:59
  • $\begingroup$ There is no work done by static friction. Only after the portion d the friction is static. $\endgroup$
    – nasu
    Feb 22, 2022 at 23:06
  • $\begingroup$ Then how would the object start rolling? Isn't the kinetic friction applied towards +x where +x is the right side? $\endgroup$
    – KayB
    Feb 23, 2022 at 15:55
  • $\begingroup$ No, the friction is to the left. Its effect is to slow down the translation and accelerate the spinning motion, until the conditon for rolling without slipping is satisfied. This happens over the distance d. The rotation is clockwise as you look from this side, isn'it? $\endgroup$
    – nasu
    Feb 23, 2022 at 16:46

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