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In case of pure rolling bottommost point is the instantaneous centre of zero velocity.

how do you $$prove\quad$$ this

Note by Manish Bhargao
2 years, 2 months ago

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This concept is called "instant center of rotation". Le me explain it mathematically (since you demand a "proof" here). Suppose a perfectly round body is undergoing a combined rotational and translational motion on a surface. Now if instead of the ground frame we observe the body from the frame of its center of mass, we will see it in pure rotation with an angular velocity, say $$\omega$$. Now the center of mass of the body is itself moving in the ground frame with some linear velocity, say $$\overrightarrow { { v }_{ o } }$$. The body can be thought of as an aggregation of n particles, with the i-th partice having mass $${ m }_{ i }$$, and velocities $$\overrightarrow { { v }_{ i } }$$ and $$\overrightarrow { { v }_{ i,cm } }$$ with respect to the ground frame and the center of mass frame of the body respectively. Then clearly $\\ \overrightarrow { { v }_{ i,cm } } +\overrightarrow { { v }_{ o } } =\overrightarrow { { v }_{ i } }$Thus the kinetic energy of the i-th particle in the ground frame is:$\frac { 1 }{ 2 } { { m }_{ i }v }_{ i }^{ 2 }=\frac { 1 }{ 2 } { m }_{ i }\left( \overrightarrow { { v }_{ i,cm } } +\overrightarrow { { v }_{ o } } \right) .\left( \overrightarrow { { v }_{ i,cm } } +\overrightarrow { { v }_{ o } } \right) =\frac { 1 }{ 2 } { { m }_{ i }v }_{ i,cm }^{ 2 }+\frac { 1 }{ 2 } { { m }_{ i }v }_{ 0 }^{ 2 }+\frac { 1 }{ 2 } { { m }_{ i }\left( 2\overrightarrow { { v }_{ i,cm } } .\overrightarrow { { v }_{ o } } \right) }$ Therefore, the total kinetic energy of the body in the ground frame is:$\\ K=\sum _{ i=1 }^{ n }{ \frac { 1 }{ 2 } { { m }_{ i }v }_{ i }^{ 2 } } =\frac { 1 }{ 2 } \sum _{ i=1 }^{ n }{ { { m }_{ i }v }_{ i,cm }^{ 2 } } +\frac { 1 }{ 2 } \sum _{ i=1 }^{ n }{ { { m }_{ i }v }_{ 0 }^{ 2 } } +\left( { \sum _{ i=1 }^{ n }{ { m }_{ i }\overrightarrow { { v }_{ i,cm } } } } \right) .\overrightarrow { { v }_{ o } }$ Now the $$\frac { 1 }{ 2 } \sum _{ i=1 }^{ n }{ { { m }_{ i }v }_{ i,cm }^{ 2 } }$$, which is the kinetic energy of the body in the center of mass frame is due to pure rotation of the body in this frame and is equal to $$\frac { 1 }{ 2 } { I }_{ cm }{ \omega }^{ 2 }$$, the rotational kinetic energy, $${ I }_{ cm }$$ being the moment of inertia of the body about its center of mass. Also $\sum _{ i=1 }^{ n }{ { m }_{ i }\overrightarrow { { v }_{ i,cm } } } =\overrightarrow { 0 }$ as this is the linear momentum of center of mass in the center of mass frame. Thus $\\ K=\frac { 1 }{ 2 } \left( { I }_{ cm }{ \omega }^{ 2 }+M{ v }_{ 0 }^{ 2 } \right)$ Now, for pure rolling, that is rolling without slipping, that is where one full rotation means travelling a length equal to the body's circumference or that is where $${ v }_{ o }=R\omega$$, we have $\\ K=\frac { 1 }{ 2 } \left( { I }_{ cm }+M{ R }^{ 2 } \right) { \omega }^{ 2 }$ where $${ I }_{ cm }+M{ R }^{ 2 }$$ is moment of inertia of the body about the point of contact with the ground, according to the parallel axes theorem. Thus the above expression for kinetic energy gives the rotational kinetic energy of the body if it is assumed to rotate purely about its point of contact with the ground at that instant. This gives a nice way of interpreting the rolling of a body on a surface without slipping. At any instant such a rolling body can now be considered to be in pure rotation about an axis through its point of contact. The point of contact, which can then be considered to be the center of rotation, can be considered to be instantaneously at rest. · 1 year, 10 months ago