NEET-XII-Physics

07: Circular Motion

with Solutions - page 5
Qstn# iv-21-a Prvs-QstnNext-Qstn
  • #21-a
    What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip?
    Ans : Given:
    Mass of the block = m
    Friction coefficient between the ruler and the block = μ
    Let the maximum angular speed be ω1 for which the block does not slip
    Now, for the uniform circular motion in the horizontal plane, we have:
    `` \mu mg=m{{\omega }_{1}}^{2}L``
    `` \therefore {\omega }_{1}=\sqrt{\frac{\mu g}{L}}``
    (b) Let the block slip at an angular speed ω2.
    For the uniformly accelerated circular motion, we have:
    `` \mu mg=\sqrt{{\left(m{\omega }_{2}^{2}L\right)}^{2}+{\left(mL{\alpha }^{2}\right)}^{2}}``
    `` \Rightarrow {\omega }_{2}^{4}+{\alpha }^{2}=\frac{{\mu }^{2}{g}^{2}}{{L}^{2}}``
    `` \Rightarrow {\omega }_{2}={\left[{\left(\frac{\mu g}{L}\right)}^{2}-{\alpha }^{2}\right]}^{1/4}``
    Page No 115: (b) Let the block slip at an angular speed ω2.
    For the uniformly accelerated circular motion, we have:
    `` \mu mg=\sqrt{{\left(m{\omega }_{2}^{2}L\right)}^{2}+{\left(mL{\alpha }^{2}\right)}^{2}}``
    `` \Rightarrow {\omega }_{2}^{4}+{\alpha }^{2}=\frac{{\mu }^{2}{g}^{2}}{{L}^{2}}``
    `` \Rightarrow {\omega }_{2}={\left[{\left(\frac{\mu g}{L}\right)}^{2}-{\alpha }^{2}\right]}^{1/4}``
    Page No 115:
  • #21-b
    If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip?
    Ans : Let the block slip at an angular speed ω2.
    For the uniformly accelerated circular motion, we have:
    `` \mu mg=\sqrt{{\left(m{\omega }_{2}^{2}L\right)}^{2}+{\left(mL{\alpha }^{2}\right)}^{2}}``
    `` \Rightarrow {\omega }_{2}^{4}+{\alpha }^{2}=\frac{{\mu }^{2}{g}^{2}}{{L}^{2}}``
    `` \Rightarrow {\omega }_{2}={\left[{\left(\frac{\mu g}{L}\right)}^{2}-{\alpha }^{2}\right]}^{1/4}``
    Page No 115: