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quawa99
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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 said:Perhaps start by asking what is the velocity of the point of contact? Is that constant or changing?
srijag said:The velocity of the point of contact during rolling motion is zero.
Tanya Sharma said:Hi quawa99...
If the velocity of the CM is vcm,radius R and the angular speed of the sphere is ω ,then what is the relationship between considering the sphere is undergoing pure rolling ?
Tanya Sharma said:vcm is with respect to the ground frame .
Anyways, now what is the relation between the acceleration of CM and the angular acceleration ?
Tanya Sharma said:Now ,what is the value of acm in the given problem ?
CWatters said:Perhaps start by asking what is the velocity of the point of contact? Is that constant or changing?
srijag said: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 said:Its not a problem given in a book
Angular acceleration and linear acceleration of cm in horizontal direction is 0
Tanya Sharma said:Since vcm = constant ,acm = 0 .
apoint of contact = apoint of contact w.r.t CM + acm
So,apoint of contact = ?
srijag said:You have to derive that. Also, acceleration of centre of mass is zero only in a uniform pure rolling.
quawa99 said:What is the acceleration of point of contact with respect to cm?
Tanya Sharma said:apoint of contact w.r.t CM = αR ,where α is the angular acceleration
quawa99 said:α is zero in my question
quawa99 said:α is zero in my question
srijag said:The equation does NOT show that acceleration is zero. It is Rw^2. Since V=Rw, it is equal to v^2/R
Tanya Sharma said: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 said: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
quawa99 said: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
quawa99 said: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
The acceleration of point of contact refers to the rate of change of velocity of a point where two objects are in contact with each other. It is a vector quantity that describes the change in the direction and magnitude of the velocity at the point of contact.
The acceleration of point of contact can be calculated using the formula a = (v2-v1)/t, where a is the acceleration, v2 is the final velocity, v1 is the initial velocity, and t is the time interval.
The acceleration of point of contact is affected by the mass, velocity, and angle of collision between the two objects in contact. The surface properties of the objects, such as friction, also play a role in determining the acceleration.
According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In the case of point of contact, the net force is the force of impact between the two objects, and the mass is the combined mass of the objects. Therefore, the acceleration of point of contact is directly related to the force of impact and inversely related to the combined mass of the objects.
Some real-life examples of acceleration of point of contact include a ball bouncing off the ground, a car colliding with another car, and a person jumping off a diving board into a pool. In all of these scenarios, there is a change in the velocity of the objects at the point of contact, resulting in acceleration.