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Rotational Mechanics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsRotational Mechanics3 Marks
Question
A solid sphere is set into motion on a rough horizontal surface with a linear speed v in the forward direction and an angular speed $\frac{\text{v}}{\text{R}}$ in the anticlockwise direction as shown in figure. Find the linear speed of the sphere:
  1. When it stops rotating.
  2. When slipping finally ceases and pure rolling starts.
✓

Answer


  1. If we take moment at A then external torque will be zero.
Therefore, the initial angular momentum = the angular momentum after rotation stops (i.e. only leniar velocity exits)
$\text{Mv}\times\text{R}-\ell\omega=\text{Mv}_0\times\text{R}$
$\Rightarrow\text{MvR}-\frac{2}{5}\times\frac{\text{MR}^2\text{V}}{\text{R}}=\text{Mv}_0\text{R}$
$\Rightarrow\text{v}_0=\frac{\text{3V}}{5}$
  1. Again, after some time pure rolling starts
Therefore,
$\Rightarrow\text{M}\times\text{v}_0\times\text{R}=\Big(\frac{2}{5}\Big) \text{MR}^2\times\Big(\frac{\text{V}'}{\text{R}}\Big)+\text{Mv}'\text{R}$
$\Rightarrow\text{m}\times\Big(\frac{\text{3V}}{5}\Big)\times\text{R}=\Big(\frac{2}{5}\Big)\text{Mv}'\text{R}+\text{Mv}'\text{R}$
$\Rightarrow\text{V}'=\frac{3\text{V}}{7}$

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