
A uniform ring of radius \[R\] is given a backspin of angular velocity \[{V_0}/2R\] and thrown on a horizontal rough surface with the velocity of the centre to be \[{V_0}\]. Find the velocity of the centre of the ring when it starts pure rolling.
Answer
548.7k+ views
Hint: Use the formula for the angular momentum of an object in terms of the linear velocity. Use the formula for angular velocity of an object in terms of angular velocity. Use the law of conservation of angular momentum. Apply this law of conservation of angular momentum to the ring performing rotational motion and its centre performing translational motion.
Formulae used:
The angular momentum \[L\] of an object is
\[L = mvr\] …… (1)
Here, \[m\] is the mass of the object, \[v\] is the velocity of the object and \[r\] is the radius of the circular path.
The angular momentum \[L\] of an object is
\[L = m{r^2}\omega \] …… (2)
Here, \[m\] is the mass of the object, \[\omega \] is the angular velocity of the object and \[r\] is the radius of the circular path.
Complete step by step answer:
We have given that the radius of the ring is \[R\]. The initial angular velocity of the ring is.
\[{\omega _i} = - \dfrac{{{V_0}}}{{2R}}\]
The initial linear velocity of the ring and its centre is \[v\].
\[{v_i} = {V_0}\]
We have asked to calculate the final velocity of the centre when the ring starts purely rotating.
Let \[m\] be the mass of the ring and \[v\] be the final velocity of the centre when its starts pure rolling.According to the law of conservation of the angular momentum, the initial angular momentum of the ring and its centre and the final angular momentum of the ring and its centre are the same.
\[{L_{Ri}} + {L_{Ci}} = {L_{Rf}} + {L_{Cf}}\]
\[ \Rightarrow - m{R^2}{\omega _i} + m{V_0}R = mvR + mvR\]
\[ \Rightarrow - \dfrac{{{V_0}}}{2} + {V_0} = 2v\]
\[ \Rightarrow \dfrac{{{V_0}}}{2} = 2v\]
\[ \therefore v = \dfrac{{{V_0}}}{4}\]
Hence, the velocity of the centre after it starts pure rolling is \[\dfrac{{{V_0}}}{4}\].
Note: The students should not forget to use the negative sign for the initial angular velocity of the ring because the ring is given a backspin angular velocity which results in the clockwise motion of the ring in the forward direction. One can also solve the same question by using the first kinematic equation for the translational and rotational motion of the ring.
Formulae used:
The angular momentum \[L\] of an object is
\[L = mvr\] …… (1)
Here, \[m\] is the mass of the object, \[v\] is the velocity of the object and \[r\] is the radius of the circular path.
The angular momentum \[L\] of an object is
\[L = m{r^2}\omega \] …… (2)
Here, \[m\] is the mass of the object, \[\omega \] is the angular velocity of the object and \[r\] is the radius of the circular path.
Complete step by step answer:
We have given that the radius of the ring is \[R\]. The initial angular velocity of the ring is.
\[{\omega _i} = - \dfrac{{{V_0}}}{{2R}}\]
The initial linear velocity of the ring and its centre is \[v\].
\[{v_i} = {V_0}\]
We have asked to calculate the final velocity of the centre when the ring starts purely rotating.
Let \[m\] be the mass of the ring and \[v\] be the final velocity of the centre when its starts pure rolling.According to the law of conservation of the angular momentum, the initial angular momentum of the ring and its centre and the final angular momentum of the ring and its centre are the same.
\[{L_{Ri}} + {L_{Ci}} = {L_{Rf}} + {L_{Cf}}\]
\[ \Rightarrow - m{R^2}{\omega _i} + m{V_0}R = mvR + mvR\]
\[ \Rightarrow - \dfrac{{{V_0}}}{2} + {V_0} = 2v\]
\[ \Rightarrow \dfrac{{{V_0}}}{2} = 2v\]
\[ \therefore v = \dfrac{{{V_0}}}{4}\]
Hence, the velocity of the centre after it starts pure rolling is \[\dfrac{{{V_0}}}{4}\].
Note: The students should not forget to use the negative sign for the initial angular velocity of the ring because the ring is given a backspin angular velocity which results in the clockwise motion of the ring in the forward direction. One can also solve the same question by using the first kinematic equation for the translational and rotational motion of the ring.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

