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QJEEM13Phe484: JEE Main 2013 Physics - Class 11 JEEM13Phe484 Solution: A hoop of radius and mass rotating with an angular velocity is…

A hoop of radius rr and mass mm rotating with an angular velocity ω0{\omega _0} is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it cases to slip?

Multiple Choice Options

  • A rω04{{r{\omega _0}} \over 4}
  • B rω03{{r{\omega _0}} \over 3}
  • C rω02{{r{\omega _0}} \over 2}
  • D rω0{r{\omega _0}}

Step-by-Step Verified Solution

JEE Main 2013 (Offline) Physics - Rotational Motion Question 203 English Explanation
From conservation of angular momentum at point of contact,

mr2ω0=mvr+mr2ωm{r^2}{\omega _0} = mvr + m{r^2}\omega

mr2ω0=mvr+mr2(vr)m{r^2}{\omega _0} = mvr + m{r^2}\left( {{v \over r}} \right) [ as v=rωv = r\omega ]

mr2ω0=mvr+mvrm{r^2}{\omega _0} = mvr + mvr

mr2ω0=2mvrm{r^2}{\omega _0} = 2mvr

v=ω0r2v = {{{\omega _0}r} \over 2}

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