- #1
AakashPandita
- 157
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Everybody says that ω=v/r. But isn't that v tangent to the cirle?
Vcom, or the velocity of the center of mass, is directly related to ω, or the angular velocity, in rolling motion. In fact, Vcom = R * ω, where R is the radius of the rolling object. This means that the linear velocity of the center of mass is equal to the product of the angular velocity and the radius.
Yes, Vcom does change during rolling motion. As the object rolls, the point of contact with the surface moves, causing the center of mass to also move and change its linear velocity. However, the angular velocity ω remains constant as long as there is no external torque acting on the object.
Vcom affects the motion of a rolling object by determining its linear velocity. The greater the Vcom, the faster the object will move. This is because Vcom is directly related to the speed at which the object rotates, which in turn affects its linear motion.
No, Vcom is not the same as the tangential velocity of a point on the rolling object. Vcom refers to the velocity of the center of mass, while the tangential velocity of a point on the object is the linear velocity of that specific point as it moves in a circular path around the center of mass.
Vcom is directly related to kinetic energy in rolling motion. This is because kinetic energy is a measure of an object's motion, and Vcom determines the linear motion of the object. Therefore, the greater the Vcom, the more kinetic energy the object will have.