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A uniform solid circular cylinder of radius r is placed on a rough horizontal surface and given a linear velocity v= 2w_0r and angular velocity w_0 as shown in the figure. The speed of cylinder when it starts rolling

  • Option 1)

    5/2 w_0R

  • Option 2)

    3/2 w_0R

  • Option 3)

    5/3 w_0R

  • Option 4)

    2/3 w_0R

 

Answers (1)

As we learnt in

Law of conservation of angular moment -

\vec{\tau }= \frac{\vec{dL}}

- wherein

If net torque is zero

i.e. \frac{\vec{dL}}= 0

\vec{L}= constant

angular momentum is conserved only when external torque is zero .

 

L_{i}=MV_{c}R+I\omega

Whence V_{c}=0

L_{i}=\frac{1}{2}MR^{2}\omega

L_{f}=MV_{c}R+I \omega' \:\:\left [\omega'=V_{c} /R\right ]

L_{f}=MV_{c}R+\frac{1}{2}MR^{2}\frac{V_{c}}{R}=\frac{3}{2}MV_{c}R

L_{i}=L_{f}

\frac{3}{2}MV_{c}R=\frac{MR^{2}\omega}{2} \Rightarrow V_{c}=\frac{R \omega}{3}


Option 1)

5/2 w_0R

This solution is incorrect 

Option 2)

3/2 w_0R

This solution is incorrect 

Option 3)

5/3 w_0R

This solution is correct 

Option 4)

2/3 w_0R

This solution is incorrect 

Posted by

Vakul

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