Acceleration of point of contact

[quawa99]

For a sphere undergoing pure rolling on a smooth surface with center of mass having constant velocity,what is the acceleration of point of contact

 

[CWatters]

Perhaps start by asking what is the velocity of the point of contact? Is that constant or changing?

 

[quawa99]

Perhaps start by asking what is the velocity of the point of contact? Is that constant or changing?

Is'n it stationary?

 

[srijag]

The velocity of the point of contact during rolling motion is zero.

 

[quawa99]

The velocity of the point of contact during rolling motion is zero.

My question is about its acceleration

 

[Tanya Sharma]

Hi quawa99...

If the velocity of the CM is v_cm,radius R and the angular speed of the sphere is ω ,then what is the relationship between them considering the sphere is undergoing pure rolling ?

 

[quawa99]

Hi quawa99...

If the velocity of the CM is v_cm,radius R and the angular speed of the sphere is ω ,then what is the relationship between considering the sphere is undergoing pure rolling ?

The velocity of of CM with respect to the point of contact is given by Vcm=RXw where w is angular velocity of the sphere.

 

[Tanya Sharma]

v_cm is with respect to the ground frame .

Anyways, now what is the relation between the acceleration of CM and the angular acceleration ?

 

[quawa99]

v_cm is with respect to the ground frame .

Anyways, now what is the relation between the acceleration of CM and the angular acceleration ?

a=RXangular acceleration

 

[Tanya Sharma]

Now ,what is the value of a_cm in the given problem ?

 

[quawa99]

Now ,what is the value of a_cm in the given problem ?

Its not a problem given in a book
Angular acceleration and linear acceleration of cm in horizontal direction is 0

 

[srijag]

Perhaps start by asking what is the velocity of the point of contact? Is that constant or changing?

I was referring to that question.

Anyway, the point of contact (in fact, all the points on the surface) have acceleration toward the centre of mass of the object. Be it, uniform pure rolling or accelerated pure rolling, the acceleration will be towards the centre of mass.
Also even if there is friction, since there is no slipping between the surfaces, the work done by it will be zero.

 

[quawa99]

Anyway, the point of contact (in fact, all the points on the surface) have acceleration toward the centre of mass of the object. Be it, uniform pure rolling or accelerated pure rolling, the acceleration will be towards the centre of mass.
Also even if there is friction, since there is no slipping between the surfaces, the work done by it will be zero.

Why do all the points on surface have an acceleration towards centre ?I have read that pure rolling is the same thing as pure rotation about point of contact so doesn't that mean acceleration of cm with respect to point of contact is v^2/R ?

 

[srijag]

Yes. Pure rolling is same as rotation about point of contact.
About the point of contact and the acceleration, mathematically you can derive for any point on the surface. And if you substitute the condition for the pioint of contact, you'll find out that the acceleration of point of contact is towards the centre.

 

[Tanya Sharma]

Its not a problem given in a book
Angular acceleration and linear acceleration of cm in horizontal direction is 0

Since v_cm = constant ,a_cm = 0 .

a_{point of contact} = a_{point of contact w.r.t CM} + a_cm

So,a_{point of contact} = ?

 

[srijag]

You have to derive that. Also, acceleration of centre of mass is zero only in a uniform pure rolling.

 

[quawa99]

Since v_cm = constant ,a_cm = 0 .

a_{point of contact} = a_{point of contact w.r.t CM} + a_cm

So,a_{point of contact} = ?

What is the acceleration of point of contact with respect to cm?

 

[quawa99]

You have to derive that. Also, acceleration of centre of mass is zero only in a uniform pure rolling.

My question is about a uniformly rolling sphere

 

[Tanya Sharma]

What is the acceleration of point of contact with respect to cm?

a_{point of contact w.r.t CM} = αR ,where α is the angular acceleration

 

[quawa99]

a_{point of contact w.r.t CM} = αR ,where α is the angular acceleration

α is zero in my question

 

[srijag]

if you consider the horizontal to be the x-axis, then, the point of contact makes an angle 3π/2. Now
a(point of contact)= a(cm)i + a(P,cm)
= a(cm)i + a(P, tangential) -i + a(P, radial)
= a(cm)i - Rα(cm) + R(w^2)
α+ angular acceleration of centre mass.

 

[Tanya Sharma]

α is zero in my question

So now what is the problem ? You have everything to answer the question .

 

[quawa99]

α is zero in my question

according to your equation the point of contact has 0 zero acceleration but then shouldn't that mean if you observe the cm in the frame of point of contact(which is stationary in present situation) it would have a centripetal acceleration of v^2/R as it is undergoing pure rotation about point of contact with angular velocity v/R

 

[srijag]

a(point of contact) = a(point of contact w.r.t CM) + a(cm)
= a(p, tangential) + a(P, radial) + a(cm) i
Since the rolling is uniform, a(cm) is zero as well as α(cm) i.e; angular acceleration of centre of mass and since a(P, tangential) = Rα(cm), it becomes zero.
Therefore, a(point of contact) = a(P, radial) = R(w^2) j
i is positive x-axis and j is positive y-axis.

 

[srijag]

The equation does NOT show that acceleration is zero. It is Rw^2. Since V=Rw, it is equal to v^2/R

 

[quawa99]

The equation does NOT show that acceleration is zero. It is Rw^2. Since V=Rw, it is equal to v^2/R

I said that for Tanya's equation not yours.I think I understood what you said. thanks :)

 

[Tanya Sharma]

quawa99..You are mixing up things ...The linear acceleration of the point of contact is zero , but it does have a centripetal acceleration about the CM owing to its rotation about the CM .

 

[quawa99]

quawa99..You are mixing up things ...The linear acceleration of the point of contact is zero , but it does have a centripetal acceleration about the CM owing to its rotation about the CM .

Isn't centrepetal acceleration also a type of acceleration?.so when you calculate the acceleration of point of contract shouldn't you think about the centripetal acceleration as well?.I never asked for linear acceleration of the point of contact specifically

 

[Tanya Sharma]

Isn't centrepetal acceleration also a type of acceleration?.so when you calculate the acceleration of point of contract shouldn't you think about the centripetal acceleration as well?.I never asked for linear acceleration of the point of contact specifically

Yes...centripetal acceleration is part of the total acceleration of the point of contact with respect to the CM .

But your original question is :

For a sphere undergoing pure rolling on a smooth surface with center of mass having constant velocity,what is the acceleration of point of contact

It doesn't ask about the total acceleration of the point of contact with respect to the CM . Generally ,if the question asks about the acceleration of the point of contact ,the implicit meaning is to calculate the linear acceleration (horizontal in this case) of the point of contact of the rolling body with respect to the ground frame.

 

[CWatters]

Certainly the original question isn't clear. When I drove my car to the shops today the point of contact certainly went with it and didn't remain in the garage. It had the same velocity and acceleration as the car.