Question

A circular disc of mass M and radius R is in pure rolling
on a horizontal surface as shown in figure. The velocity
of its centre of mass is Vo. Then the ratio of net velocity
of point A and point B will be​

Answer 1

Given:

A circular disc of mass M and radius R is in pure rolling on a horizontal surface as shown in figure. The velocity of its centre of mass is Vo.

To find:

Net velocity at point A and B ?

Calculation:

Since the disc is performing pure rolling, at the point of contact we can say :

$V_{0} = \omega R$

So, at point B , net velocity will be :

$V_{net} =V_{0} - \omega R$

$\implies V_{net} =V_{0} - V_{0}$

$\implies V_{net} =0$

Now, at point A, we can say:

$V_{net} = \sqrt{{(V_{0})}^{2} + {(\omega R)}^{2} }$

$\implies V_{net} = \sqrt{{(V_{0} \: )}^{2} + {(V_{0})}^{2} }$

$\implies V_{net} = \sqrt{{2(V_{0} \: )}^{2} }$

$\implies V_{net} =V_{0} \sqrt{2}$

Hope It Helps.

Answer 2

Explanation:

Given:

A circular disc of mass M and radius R is in pure rolling on a horizontal surface as shown in figure. The velocity of its centre of mass is Vo.

To find:

Net velocity at point A and B ?

Calculation:

Since the disc is performing pure rolling, at the point of contact we can say :

V_{0} = \omega RV

0

=ωR

So, at point B , net velocity will be :

V_{net} =V_{0} - \omega RV

net

=V

0

−ωR

\implies V_{net} =V_{0} - V_{0} ⟹V

net

=V

0

−V

0

\implies V_{net} =0⟹V

net

=0

Now, at point A, we can say:

V_{net} = \sqrt{{(V_{0})}^{2} + {(\omega R)}^{2} }V

net

=

(V

0

)

2

+(ωR)

2

\implies V_{net} = \sqrt{{(V_{0} \: )}^{2} + {(V_{0})}^{2} }⟹V

net

=

(V

0

)

2

+(V

0

)

2

\implies V_{net} = \sqrt{{2(V_{0} \: )}^{2} }⟹V

net

=

2(V

0

)

2

\implies V_{net} =V_{0} \sqrt{2} ⟹V

net

=V

0

2

Hope It Helps.