A uniform solid cylinder is given an angular speed and placed on a rough plate of negligible thickness. The horizontal surface below the plate is smooth. Then the angular speed of the cylinder when it starts pure rolling on the plate will be: [ Assume sufficient length of plate ]
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please see attached file
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Since friction is acting at the lower most point P, angular momentum conservation can be applied about the lower most point.
Angular momentum initial
Li=MVcmR+IwLi=MVcmR+Iw
=Iw=Iw[since vcm=0vcm=0]
=MR2w2=MR2w2
Angular momentum final
Lf=MVcmR+Iw′Lf=MVcmR+Iw′
w′=VcmRw′=VcmR
∴Lf=MVcmR+MR22VcmR∴Lf=MVcmR+MR22VcmR
=32=32MVcmRMVcmR
Li=LfLi=Lf
3232MVcmR=MR2w2MVcmR=MR2w2
Vcm=Rw3
Angular momentum initial
Li=MVcmR+IwLi=MVcmR+Iw
=Iw=Iw[since vcm=0vcm=0]
=MR2w2=MR2w2
Angular momentum final
Lf=MVcmR+Iw′Lf=MVcmR+Iw′
w′=VcmRw′=VcmR
∴Lf=MVcmR+MR22VcmR∴Lf=MVcmR+MR22VcmR
=32=32MVcmRMVcmR
Li=LfLi=Lf
3232MVcmR=MR2w2MVcmR=MR2w2
Vcm=Rw3
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